NE and SOLID
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i ADVANCED STUDENTS
SON and BAXANDALL
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The University of Toronto
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PRACTICAL
PLANE AND SOLID GEOMETRY
PRACTICAL
PLANE AND SOLID
GEOMETRY
FOR ADVANCED STUDENTS
INCLUDING GRAPHIC STATICS
ADAPTED TO THE REQUIREMENTS OF THE
ADVANCED STAGE OF THE SOUTH
KENSINGTON SYLLABUS
BY
JOSEPH HARRISON
M.I.M.E. , ASSOC. M. INST. C.E.,
INSTRUCTOR IN .MECHANICS AND MATHEMATICS AT THE ROYAL
COLLEGE OF SCIENCE, LONDON
AND
G. A. BAXANDALL
ASSISTANT IN MECHANICS AND MATHEMATICS AT THE ROYAL
COLLEGE OF SCIENCE, LONDON
Eonbott
MACMILLAN AND CO., Limited
NEW YORK : THE MACMILLAN COMPANY
l8 99
an
tfZl
PREFACE
The training in Practical Geometry suitable for the student
of Art consists of working problems mainly in Plane
Geometry and Perspective. The former because decora-
tive designs are largely based on geometrical figures ; and
the latter because an artist constantly deals with appearances.
On the other hand, the student of Science finds in
Practical Geometry a powerful engine of calculation, which
may often with great advantage be employed in preference
to analytical processes. The principal use to him of plane
geometry is that it enables him to make graphical or
semi-graphical computations. Descriptive geometry is of
great importance since it is required in problems which
involve three dimensions in space ; and also because it
shows how to exhibit the actual forms and dimensions of
solid objects.
This book is written for Science students. The
necessity of accurate draughtsmanship is insisted on
throughout. We describe how the drawing instruments
may be set and their efficiency maintained. And the
numerical answers appended to many of the examples
should help to prevent any relapse into slovenly and in-
accurate work on the part of those students who are apt to
become lax.
In Euclid's system of pure geometry the mechanical
VI PRACTICAL GEOMETRY
appliances and geometrical constructions allowed are
severely restricted, the former being limited to a straight-
edge, pencil, compasses, and a single plane surface on which
to draw. A length may be transferred or a circle drawn by
using the compasses, and a straight line may be drawn
through two points. A line may not, however, be drawn to
touch two circles without first finding the points of con-
tact, although this adds nothing to the practical accuracy of
the result.
These restrictions, which form an important part of
Euclid's scheme of logic, may yet be quite unsuitable in
actual drawing. And in fact, if strictly adhered to, they
are found to hamper the student in his work, to be wasteful
of time, to lead to inaccuracies which may nullify the result,
and even to render problems impossible of solution, the
difficulties of which may otherwise be successfully over-
come. In illustration see Example 2, page 139, on the
motion of a slide valve driven by a Gooch link.
It is true that in examinations in Practical Geometry
the use of the tee- and set-squares has caused Euclid's
restrictions to be somewhat relaxed, in that parallel and
perpendicular lines are usually permitted to be drawn
without construction. But does this go far enough ? Is
not our science still in leading-strings, tied to the systems
of the pure geometrician, its humble function being to
illustrate his principles ?
No doubt a student already familiar with the proofs of
Euclid may derive much benefit from illustrating the more
important propositions by carefully drawing the figures to
scale. For example he may learn how small need be the
errors which are introduced into graphical work. At various
parts of Section I. such examples have been inserted.
PREFACE vii
But in the authors' view a fuller conception of the scope
and province of Practical Geometry is needed, one which,
seeing in it a powerful instrument of mathematical investi-
gation, recognises the limitations as to accuracy and the
practical requirements of the draughtsman, and permits
and teaches the use of any mechanical devices where these
are found to be useful and beneficial as, for example, the
employment of transparent templates, to be adjusted by
trial by hand, such as are described in connection with
some of the problems and examples of this book.
Chapter VI., which deals with the use of squared paper,
and the plotting of curves from co-ordinates, should be
found useful in view of the increasing recognition of the
importance of this branch of the subject.
In Chapter VII., where three planes of reference are
used, the treatment is somewhat new. But the method
has been tried and interests students, and is introduced
with confidence. It is intended that a student shall read
this chapter without much assistance from his teacher.
Thus at the beginning of Descriptive Geometry he gets
some very clear notions of projection and of the geometry
of space, and is also trained to improvise models where
these are helpful.
The chapter on metric projection goes rather beyond
the requirements of the advanced stage. But to have
omitted any part would have left the subject very
incomplete.
The space at our disposal for the section on Graphics
was very limited, but it is hoped that this part has not
been unduly condensed.
It is no doubt convenient to have the figures and
corresponding descriptions arranged so as to avoid the
viii PRACTICAL GEOMETRY
necessity of turning over a leaf when referring to a
diagram, but occasionally this could only be done at the
expense of crowding or condensation.
Examples bearing on the problems will generally be
found following the latter or in close proximity thereto.
The miscellaneous examples which close the chapters are
mainly derived from the past examination papers of the
Department of Science and Art, advanced stage, and by
working these the student will be able from time to time
to test his proficiency.
In order to ensure the attainment of the desirable degree
of accuracy in graphical work generally, the figures should
not be made too small. If in some cases it may appear that
this condition has been ignored, the reason must be attri-
buted to want of space. The student, with drawing-paper
at his disposal, cannot urge this excuse.
July 1899.
CONTENTS
SECTION I. PRACTICAL PLANE GEOMETRY
CHAPTER I
hAGE
General Introduction . , i
CHAPTER II
Similar Rectilineal Figures Areas . , 26
CHAPTER III
Triangles Circles and Lines in Contact . 50
CHAPTER IV
Conic Sections .... .70
CHAPTER V
Special Curves .... .110
CHAPTER VI
Co-ordinates Plotting on Squared Paper . 148
PRACTICAL GEOMETRY
SECTION II. PRACTICAL SOLID GEOMETRY,
OR DESCRIPTIVE GEOMETRY
CHAPTER VII
PAGE
Position in Space defined and exhibited . . 176
CHAPTER VIII
Fundamental Rules ok Projection . . . 193
CHAPTER IX
The Straight Line and the Perpendicular Plane . 204
CHAPTER X
The Oblique Plank ..... 230
CHAPTER XI
Horizontal Projection, or Figured Plans . . 274
CHAPTER XII
Plane and Solid Figurks in given Positions . . 292
CHAPTER XIII
The Projection of Curves and Curved Surfaces . 308
CHAPTER XIV
Tangent Planes to Surfaces . . . 34
CONTENTS xi
CHAPTER XV
PAGE
Surfaces in Contact ..... 368
CHAPTER XVI
Intersections of Surfaces, or Interpenetkations of
Solids ....... 384
CHAPTER XVII
Cast Shadows ...... 406
CHAPTER XVIII
Metric Projection . . . . 442
CHAPTER XIX
Miscellaneous Problems . 462
SECTION III. GRAPHICS
CHAPTER XX
Graphic Arithmetic ..... 4S6
CHAPTER XXI
Graphic Statics . , . 502
xil PRACTICAL GEOMETRY
APPENDIX I
PAGE
Science and Art Department Examination, May 1899 538
APPENDIX II
Definitions and Theorems ok Pure Solid Geometry 542
INDEX .... . 549
SECTION I
PRACTICAL PLANE GEOMETRY
CHAPTER I
GENERAL INTRODUCTION
1. Scope of subject. Practical geometry is the science
which is concerned with applications of the problems of
pure geometry to cases in which the geometrical figures
require to be drawn to scale, often with great accuracy.
In pure geometry we have a set of propositions,
arranged in logical sequence, and deduced by strict reason-
ing based on a few fundamental axioms and certain mathe-
matical abstractions or conceptions which are defined.
Hand sketches serve the purpose of representing the ideal
figures.
In practical geometry the conditions are different. The
principal requirement is the careful drawing, to scale, of
geometrical figures by means of mathematical instruments.
More freedom is allowed in the use of instruments. Thus,
by means of the edge of the drawing-board, by tec-square
and set-squares, by parallel rulers, and by the clinograph, we
draw parallel, perpendicular, and symmetrically disposed
lines without any construction, such as would be required
by Euclid.
2 PRACTICAL PLANE GEOMETRY chap.
2. Limits of accuracy. In pure geometry a point is
defined as having position, but not magnitude, and a line
has no breadth. Now it is evident that, in order to be
visible, a point must be of a definite size, and a line must
have breadth. Moreover, it is impossible to actually draw
a line which shall be perfectly straight, or a circle, curve,
or, in fact, any figure with absolute accuracy. Consequently
the results we obtain by graphical construction are subject
to unavoidable errors, and in practical geometry these con-
ditions are recognised at the outset.
In striving to approximate as nearly as may be to the
ideal, and to obtain the greatest degree of accuracy war-
ranted by the conditions of the particular problem under
consideration, we fix working limits to the errors within
which our results should be confined.
Thus a point, pricked in the paper with the point of a
needle, need not be more than -j-j^- of an inch in diameter
in order to be readily seen by a person with ordinary eye-
sight. A line drawn with a lead pencil of suitable lead
and properly sharpened need- not be more than tt^-q- of an
inch wide ; and with a good " straight-edge," no part of a
line, of say one foot in length, need be out of the true
alignment by so much as the -g-^-g- of an inch. The lines
on an engine-divided scale should be correct in position to
within xoVo f an inch, and with care we should be able to
measure and set off distances to within -g^ of an inch.
Set-squares are easily adjusted, so that the errors in the
angles are kept within one minute or -^ of a degree.
An expert draughtsman would be able to reduce these
errors, if the circumstances were such as to warrant the
increased care and expenditure of time that would be neces-
sary. The limits given above are ordinary practical work-
ing limits, and should be expected to be attained by every
student in his work. In order to maintain this standard of
accuracy, the student should have access to the tools
necessary to keep his instruments in good working order,
such as the trying-plane, oil-stone, etc.
GENERAL INTRODUCTION
3. Instruments. A list of the principal drawing instru-
ments required by the student is now given, with a brief
description of some of them.
The drawitig-board should be of "imperial" size, 30" x
22", and the tee-square of corresponding length. A "half-
imperial " board in addition is very convenient.
The 6o set-square may be 9" or 10" long, and the 45
set-square about 6" long.
A clhiograph is almost indispensable ; see next article
for illustration.
The pencils should be of pure quality, and of hard lead,
say HH, or HHH.
The compasses, if the best, will have knee-joints and
needle-points firmly secured, and the dividers will have a
fine screw adjustment. The inking-pens need not have
jointed nibs.
A pricker is requisite, and may be made by breaking
a length of about 1" from the sharp end of a stout needle,
and inserting it in the wood of a penholder, leaving V or
so of the point projecting.
A 6" semicircular protractor is preferable to the common
6" rectangular one, as an instrument for measuring and
setting off angles with accuracy ; but the best form is the
circular protractor, and the diameter of this may be about
6". It is desirable that angles should be capable of being
read or marked off to within -Jg- of a degree.
The scale is a very important instrument, and a special
description of it is given in Art. 5.
French curves may be used as templates when lining in
curves which have first been carefully drawn freehand.
Good cartridge drawing-paper with a smooth surface
answers every purpose. Dratving-pins if used should have
thin heads, but the latter always impede the movement of
the squares. To overcome this, a good plan is to secure
the corners of the paper with sealing-wax. The student
should avoid inserting pins in his boards anywhere but
near the edges.
PRACTICAL PLANE GEOMETRY chap.
4. Setting of instruments. The lead pencil should be
of hard quality, and finished to a fine chisel edge with glass
paper as illustrated in the figure opposite. The leads of
the compasses may be similarly sharpened.
The edges of the drawing-board and of the tee-square and
set-squares should be trued up, and the angles of the latter
made accurate by using a trying-plane.
The dinograph is useful for drawing series of lines
which are respectively parallel, perpendicular, or symmetric-
ally inclined. See Example 6.
Examples. 1. Draw a fine line, and on it step off with the
dividers ten divisions each \" long ; and from the last point
step off one division in a direction at right angles to the others.
Join the first and last points, and through the other points draw
lines parallel to this line. The lines so drawn will be nearly
T V' apart, and should be so fine as to be each distinct.
2. Repeat example I, taking the divisions each T V' instead of A".
The lines will now be only -j-JV' a P ai 't, and should still be dis-
tinct from one another.
3. By using the tee-square and set-square draw two lines cross-
ing one another at right angles. Then starting from any
point on one of the lines, and using the 45 set-square, draw
four lines in succession across the angular spaces, meeting
respectively on the lines first drawn. The last point should
coincide with the first point, the four lines last drawn forming
a square. If this figure be drawn big enough, it gives a very
severe test of the accuracy of the 45 set-square.
4. Draw a circle of about 4" radius, and through its centre draw
the six possible lines with the tee- and 6o set-squares. Prick
off the twelve points where these lines intersect the circle, and
test whether the twelve chords joining these points are of
equal lengths. Also test whether the directions of four of the
chords agree with the 45 set-square, and whether the 6o and
45 set-squares can be arranged with two of their edges in
contact, and one resting on the tee-square, so that the upper
edge of the outer square shall coincide in succession witii the
eight remaining chords.
5. Describe a circle of say ii/' radius ; draw the circumscribing
hexagon by using the tee-square and 6o set-square. Prick
off the corners of this hexagon, and test whether all the sides
are of equal length. Also try whether the nine lines joining
the alternate and opposite points agree in direction with one or
other of the edges of the 6o set-square.
GENERAL INTRODUCTION
6. Let any line XY be drawn and a point P taken anywhere on
the drawing-paper. It is required to draw through P, (a)
a line parallel to XY; (b) a line perpendicular to XY; (c) a
line symmetrically inclined with XY.
(a) Place the edge AB of the clinograph on the edge of the tee-
square ; move the tee-square along the edge of the board and
the clinograph along 'the edge of the tee-square, setting the
edge DE to coincide with XY. Again move tee-square (if
necessary) and clinograph until DE passes through P, when
the required line may be drawn.
(b) After setting the edge DE to coincide with XY, turn the
clinograph over through a right angle so that BC rests on the
edge of the tee-square ; the required line through P is then
readily drawn.
(C) Proceed as in (a) ; then turn the clinograph on to its other
face, still keeping AB on the edge of the tee-square ; a line
may then be drawn through /'such that this line and A' Fare
symmetrical with regard to horizontal and vertical lines.
6 PRACTICAL PLANE GEOMETRY chap.
5. The Scale. The most useful scales for our work are
those which are decimally subdivided. The subdivisions
may extend along the whole length, or be confined to end
main divisions, being technically known as " close " and
"open" divided scales respectively. One close divided
and two open divided scales can be got on one edge, so
that if both faces be used, four different scales of the former
kind and eight of the latter can be set off on one piece of
boxwood, by using all the available space on it.
The figure shows portions at the ends of one face of a
boxwood scale 1 2" long, with four open divided scales of 1",
|", \", and \" respectively, decimally subdivided. On the
other face it would be very convenient to have four scales
forming the set f ", ", T V, s\", or the set ", \'\ ", T V",
also decimally subdivided. The student should be pro-
vided with one or other of the set of eight scales thus
described.
The divisions of the scale are marked o, 1, 2, 3, . . ., but
they may be read as o, 10, 20, 30, . . ., or o, 100, 200,
300, . . ., etc., or as o, .1, .2, .3, . . ., or o, .or, .02,
.03, . . ., or o, .001, .002, .003, . . ., and so on.
If the divisions are read, o, 1, 2, 3, . . ., as marked,
then the subdivisions represent figures in the first decimal
place, and a fraction of a subdivision could be estimated
by a figure in the second decimal place. Thus the distance
PQ on the Y scale would be read 3.24.
If the figures marked are read as tens, i.e. 10, 20, 30,
. . ., the subdivisions become units, and the fraction of a
subdivision is in the first place of decimals. Thus the
distance PQ would now be 32.4.
If the readings of the open divisions are 100, 200, 300,
. ., PQ reads 327. If 1000, 2000, 3000, . . ., PQ is
3240, and so on.
While if the numbers at the open divisions stand for
,i, .2, .3, . . ., then PQ represents .324. If the numbers
are taken to mean .01, .02, .03, . . ., PQ is read equal to
.0324, and so on.
GENERAL INTRODUCTION
Q
J-
y
/
2
4
6
8
10
'8
o \
T
,
/
<LM.li
i 1 i 1 i 1 i 1 i
u
'ID
yz
o
^
yl
^
Examples. 1. Measure the distance AB :
(a) On the scale of i" to I unit. Ans. 2.37.
(b) On the scale of i" to ioo units. Ans. 237.
(c) On the scale of \" to 10,000 units. Ans. 47500.
(if) On the scale of |" to o. I unit. Ans. 0.95.
(e) On the scale of ' to .001 unit. /4.r. 0.00475.
(/) On the scale of ^" to 10 units. Ans. 71.3.
(g) On the scale of |" to .01 unit. Ans. 0.0317.
Note. The respective scales must be applied directly to the line,
and the number of units read off, without the interposition of dividers.
2. Mark off lengths of-
(a)
{<)
(d)
w
3.78 units on the scale of 1" to I unit.
697 units on the scale of ^" to 100 units.
0.913 unit on the scale of i" to o. 1 unit.
0.001427 unit on the scale of j" to .0001
0.092 unit on the scale of " to o. 1 unit.
unit.
Note. The lengths must be marked off with a pencil or pricker
direct from the scale, the latter being applied to the paper. Dividers
should not be used.
3. Set off lengths representing :
(a) 2.37 feet, to the scale of 1 inch to the foot.
(b) 64,500 lbs. to the scale of \ inch to 10,000 lbs.
PRACTICAL PLANE GEOMETRY chap.
6. Ratio and proportion. When we speak of the ratio
of one magnitude to another, say of A to B, written A : B
or A/B, we signify how many times the first contains the
second.
Thus comparing a guinea with a crown, their respective
money-values are in the ratio of 21 to 5, and this ratio 2 y x
equals 4.2. Thus a ratio may be expressed as a pure
number, that is, one not associated with kind or quantity.
Or again the numbers on an ordinary scale give the
ratios of the several lengths to the unit length.
Proportion implies the equality of two ratios. Let the
ratio A : B between two magnitudes be equal to the ratio
a : b between two other magnitudes, the four terms A, B, a, b,
are said to form a proportion. This equality is expressed
in symbols thus
A : B : : a : b,
or A: B = a: b,
A a
B=b>
and by algebra it is readily seen that
B h r> a 1
= --. or B : A : : b : a ;
A a
A B
also = , or A : a : : B : :
a b
and A x b = a x B.
Similar polygons afford typical illustrations of ratios and
proportion. The student is acquainted with the funda-
mental property of such figures, viz. :
Theorem. In two similar rectilinear figures the ratios of
pairs of corresponding tines are all equal.
Thus in the similar triangles ABC, Abe we have
AB BC CA
Ab " be ' cA '
or AB : BC: CA = Ab : be : cA.
GENERAL INTRODUCTION
And in the similar figures OABCD, Oabcd the ratios
AB BC CD DA OA OB OC OD
cd ' ~da J ~0a' ~0b' ~Oc'
ab ' be '
are all equal.
Od
The following relations are important. The student should verify
them by inserting simple numbers for the letters.
M N
If
then
rind
in
p-M+q-N M N
= or ,
p * m + q n m 7i
p-M+q-m p-N+q->r
'Jlf+s'tn r'J\T+s'n
where p, q, r, and s are any numbers, positive or negative, whole or
fractional. Thus in the figure,
Ab Ac AB-Ab AC -Ac bB
or
_ Ac
A~B = AC' :
AB
AC
AB'
cC
AC'
Examples. 1. Measure the sides of the triangles ABC, Abe to
three significant figures on any suitable scale, say by applying
the j" decimal scale direct to the figure ; and, by numerical
calculation, verify the equality of the three ratios stated on p. 8.
2. Verify the equality of the given ratios for the quadrilaterals by
measuring the lengths of all the lines and performing the
divisions by arithmetic. Calculate to three significant figures ;
find the average of the eight values, and taking this as the
correct value of the ratio, estimate the percentage error of the
one most out of truth.
IO
PRACTICAL PLANE GEOMETRY chap.
7. Problem. To divide a given line AB into any
number of equal parts, say seven.
(a) By construction (no figure). Draw AC at any
angle, say 45, to AB, and, selecting any unit, step it off
with the dividers seven times along A C, from A to D ;
or applying the scale to AC, mark off, by means of pencil
or pricker, seven equal divisions. Join DB, and through
the other points of division on AD draw lines parallel to
DB ; these intersect AB at the required points.
Note. In choosing the unit, care should be taken that AD is neither
too short nor too long, or the construction becomes "ill conditioned,"
and the points on AB are not determined with sufficient precision.
(/<) By trial with the dividers. This is the method of
trial and error and should be encouraged. It is a training
in the estimation of lengths, and with practice the required
division is quickly and accurately effected.
(c) By the use of ruled (or squared) tracing-paper or
ruled celluloid. Place the ruled tracing-paper as shown
over AB, with the line marked o passing through A. Put
the point of a pricker through the tracing-paper at A and
turn the paper round until the line 7 passes through B.
Then carefully prick through at the places where the lines
between o and 7 cross AB.
These ruled templates will be found very useful in graphical work,
and it is worth while for the student to make them of celluloid, which
may be done thus : Paper accurately ruled (or squared paper) may
be readily obtained, or failing this, carefully set out ; the sheet celluloid
is then placed over this and the lines scratched with a needle-point.
Two sizes of templates should be made. The smaller one may have
the twelve main lines 4" long and 0.2" apart, each division being sub-
divided into five equal parts. Allowing a margin of h" all round, the
size of the template would be 5" by 3V' outside. The larger template
might conveniently have the twelve main lines 7" long and V' apart,
each being subdivided into ten. The outside dimensions would be 8"
by 5", leaving the same margin as before.
The lines should be scratched of different widths for the sake of
clearness.
GENERAL INTRODUCTION
ii
Examples. 1. Draw a straight line, and on it mark off a length
A B of 3.67". Bisect AB.
Note. The bisection of a line is best done with the compasses,
using the lead. Set the compasses to the half-length as nearly as
can be guessed, and witli A and B as centres describe short arcs
nearly meeting on the line. A second guess may now be made which
is very nearly, if not quite, accurate, the arcs drawn, and the mid point
between them estimated.
2. Divide the line of Example I into 8 equal parts.
Note. Here the method of continual bisection is useful.
3. On a straight line mark off a length of 3.14", and divide this
length into 5 equal parts, by each of the methods of this article.
12 PRACTICAL PLANE GEOMETRY chap.
8. Problem. To divide a line AB into consecutive
parts which have given ratios to each other, say 2:3:5.
(a) Draw a line through A inclined at any angle to AB,
and to a convenient scale mark off A C, CD, DE, respect-
ively equal to 2, 3, 5 units.
Join EB and draw CT^and DG parallel to EB. Then
AF-.FG: GB = AC : CD : DE = 2 : 3 : 5 (Euclid, VI. 10).
(0) By the use of ruled tracing-paper or celluloid. This
may be effected in a manner similar to that already de-
scribed in Art. 7. The sum of 2 and 3 is 5, and that of
2, 3, and 5 is 10. So by placing the line o over A, and
the line 10 over B, the required points of division are
obtained by pricking through where the lines 2 and 5
intersect AB.
If decimals occur in the given ratios, decimal subdivi-
sions are required in the ruled template.
9. Problem. The given line P represents a speed of
24 6 feet per second ; construct the scale of speed, and
measure the speed represented hy the line Q.
(a) On any line set off RS equal to P, and draw ST
perpendicular to RS.
With centre R, radius 24.6 measured on a suitable
scale, describe an arc intersecting ST in T, and join
RT.
Set the scale used in measuring 24.6 with its edge along
RT, and mark or prick off the decimal divisions and sub-
divisions shown.
Perpendiculars from these to RS will give the required
divisions of the scale of speed. The scale is then numbered
as shown.
Measuring Q on this scale, the speed represented by
it is found to be 1 1.3 feet per second.
(/>) By the use of a ruled template. Set the template
with the zero line on R, and the 2.46 line over S, then
prick through the decimal divisions and subdivisions of the
scale.
GENERAL INTRODUCTION
13
F G B
Feet ptr second/
A
1
B
r
D
Examples. 1. Divide a line 2. 87" long into two parts which shall
be in the ratio 4:7.
2. Divide a line 3. 19" long into two parts, so that the length of
one part is to the length of the whole as 4 : 7.
3. On a straight line mark off a length of 2.71", and divide the
length into three consecutive segments, having the ratios to one
another of 2.3, 3.6, 1.7.
4. Measure the ratio of AC to AB in the bottom figure. Ans. 2.80.
Place the zero line of the template on A, and rotate until one
of the main division lines comes over B ; read off the position
of C, and divide by B.
5. Measure the ratios (a) AB-.AC : AD. Ans. 1 : 2.80 : 3.70.
\b) DC : DB : DA. Ans. 1 : 3.06 : 4.20.
(c) AC-.BD. Ans. 1.05 : 1.
I4 PRACTICAL PLANE GEOMETRY chap.
10. Problem. To construct a diagonal scale of 1J" to
the foot, which shall measure feet, inches, and eighths of
an inch.
Set off eight equal divisions along a vertical line AB,
and draw the nine horizontal lines through the points of
division.
Draw a series of vertical lines, \V' apart, for the main
divisions of the scale, representing feet.
Apply the |" scale to BC and AO, mark off twelve
equal divisions on each of these lines, and draw the twelve
diagonal lines between them.
Figure the scale as shown.
The length between the two dots represents i 2".
11. Problem. Linear measure in Western India being
as follows :
1 tasu =1125 inches
1 hath = 16 tasu = 1 foot 6 inches
1 gaz =l l hath = 2 feet 3 ,,
Draw a diagonal scale of -$ showing gaz, hath, and tasu.
Mark off on it two lengths respectively equal to 2 gaz,
hath, 12 tasu ; and 1 gaz, 1 hath, 4 tasu. (1898)
Mark off sixteen equal divisions along a vertical line
oZ>, and draw the seventeen equally spaced horizontal
lines.
By calculation, Ts V of 2' 3" = --" = 0.9 '. Therefore
draw the vertical lines of the scale at horizontal intervals of
0.9" to represent gaz.
Set off DE= 1 hath = gaz = 0.6". Draw the diagonal
line BO, and through 8 draw 81 parallel to BO.
Figure the scale as shown.
The two required distances are indicated on the scale
by dots.
Note. Diagonal scales have the disadvantage that the divisions do
not extend to the edge, and they are not more accurate in use than
ordinary scales. They are however useful where, as above, it is desired
to represent quantities which are expressed in three denominations.
GENERAL INTRODUCTION
15
B C
3A
/'l
M
f&
M
M
4
MM
M
M M M M M
& 9" 6"
tdAiV E
16
12
O
10
/'
D
..... 1
-X .,,,., -o - .. -., ,
N 1
hcUhi
zgaz
11
Examples. 1. Construct a diagonal scale, the representative
fraction of which is -fa, reading yards, feet, and inches.
2. Draw a scale showing hundredths of an inch by diagonal
division.
3. Draw a scale of feet to measure all distances between 70 feet and
1 foot, where 5| feet is represented by .52 inch ; and by
diagonal division, make this scale available for reading inches.
4. Construct a scale of 120 feet to an inch, to measure 700 feet,
from which single feet can be taken.
5. Construct a scale of 100 fathoms, with iS fathoms represented
by 1 inch, from which feet may be measured.
6. Construct a scale of 76 miles to 1.3 inches, to read to single
miles, and to exhibit 500 miles.
7. A volume of 374 cubic inches is represented by a line 4.61 inches
long. Construct a decimal scale of volume, and by its use
measure the volume represented by a line 5.6 inches long.
Am. 455 cubic inches.
16 PRACTICAL PLANE GEOMETRY chap.
12. Solution of the right-angled triangle. Many subse-
quent problems reduce to that of solving a right-angled
triangle, having given two of its elements.
The jfo<? elements (exclusive of the right angle) are
the height
a,
the base
b,
the hypothenuse
c,
the base angle
A,
the vertical angle
B.
There are nine possible cases. To solve the triangle
i. Given the height and base a, b,
2. given the base and hypothenuse b, e,
3. given the hypothenuse and height c, a.
4. Given the height and base angle a, A,
5. given the base and base angle b, A,
6. given the hypothenuse and base angle c, A.
7. Given the height and vertical angle a, B,
8. given the base and vertical angle b, B,
9. given the hypothenuse and vertical angle c, B.
The student will readily solve all these cases himself.
We have placed them here for convenience of reference,
and to direct attention to them.
Examples. Solve the following four right-angled triangles by
accurately drawn figures, and measure the results :
1 Given the hypothenuse 3.2" and the height 1.9".
Ans. b = 2.sf,A = z6.Ar\ 5 = 53-6.
2. Given the height 2.18", base angle 36. 3 .
Ans. = 2.97", 6 = 3.68", B = S3-7-
3. Given the base 3.06", vertical angle 49.1.
Ans. a = 2.66", c = 4.os", A =40.9".
4. Given the hypothenuse 3.92", vertical angle 55.7.
Ans, a -2.21", =3.24", ^ = 34.3.
GENERAL INTRODUCTION
17
5. Measure all the sides and angles of the triangle ABC, the linear
scale being j. Square the two sides a and b and add, and
compare this with the square of c.
6. The shadow cast by a vertical post 57" high on level ground was
measured and found to be 87.6" ; find the altitude of the
sun above the horizon. Ans. 33.1.
7. Wishing to know the height of an electric arc light, I placed a
5-feet rod vertically upwards on the floor, and found its shadow
to measure 4.2 feet ; on moving the rod 6.2 feet along the
floor directly away from the light its shadow measured 7.5 feet.
Required the height of the light above the floor. Ans. 14.4
feet.
8. Two knots on a plumb-line at heights of 7 feet and 2 feet above
the floor cast shadows at distances of 1 1.4 feet and 1. 65 feet re-
spectively from the point where the line meets the floor. Find the
height of the source of light above the floor. Ans. 12. 1 feet.
9. A river AC, whose breadth is 217 feet, runs at the foot of a tower
CB, which subtends an angle BAC of 27.4 at the edge of the
opposite bank. Required the height of the tower. Ans. 112.4
feet.
10. A person on the top of a tower 68 feet high observes the angles
of depression of two objects on the horizontal plane, which are
in line with the tower, to be 32. 4 and 48. 6. Find their
distance from each other and from the observer. Ans. 47.1
feet, 90.7 feet, 127 feet.
11. The hypothenuse of a right-angled triangle is 3.45 feet, and one
of the sides is double the other ; determine the sides and angles.
Ans. 1.54 feet, 3.08 feet, 26.5 , 63. 5 .
12. The hypothenuse of a right-angled triangle is 43.5 feet, and
one of the adjacent angles is double the other ; find the sides
and angles. Ans. 21.7 feet ; 37-7 f ee t-
C
18 PRACTICAL PLANE GEOMETRY chap.
13. Ways of defining angles. Let the student draw
two straight lines at random, meeting at a point, and
including some angle. By the use of any instrument
except the protractor, let him obtain some information from
the angle, which, being given to a neighbour, shall enable
the latter to construct an angle of equal size. A number
of different ways of doing this will now be given, but the
student should think one or two out for himself before
reading farther.
Definitions. Let A OB be any angle. With centre O,
and any radius, describe the arc RS, intersecting OB, OA,
in R and S, and draw the chord RS. Then it will be
quite clear that if the lengths of the radius OR and chord
RS were given, this would be sufficient information to
enable one to construct an angle equal to A OB.
Again, in OB take any point P, and draw PN per-
pendicular to OA, it is evident that a knowledge of the
lengths of ON and NP would enable the angle to be repro-
duced in size.
We might, instead, give the lengths of NP and PO, or
of ON and OP. Whatever pair we take, it will be a simple
matter to construct an angle equal to A OB, remembering
always that the angle ONP must be a right angle.
But observe carefully that it is not the actual lengths of,
say, ON and NP which are necessary, for the lengths of
ON' and NP would do equally well. What, then, is it
sufficient to give ? The answer is that we may give the
ratio of ON to' NP, or NP to OP, or ON to OP, or RS to
OS, or R'S' to O'S ; for then the student may take any
convenient lengths for the pair of lines, so long as they are
in the given ratio.
For example, if ON were taken three inches and NP
one inch, the same angle would be determined as by taking
ON' 6 inches and N'P' 2 inches, 3 to 1 being the same
ratio as 6 to 2.
It will be convenient to give names to these ratios. The
following are in constant use :
GENERAL INTRODUCTION
19
S S'A
RS
OR
NP
OP
ON
OP
NP
ON
is called the chord of the angle A OB
sine
cosine
tangent
These are abbreviated to cho AOB, sin AOB, cos AOB,
and tan AOB respectively. Thus
cho AOB =
sin AOB =
cos AOB =
tan AOB =
RS
"OR
PN
hypothenuse OP
base ON
chord
radius
height
hypothenuse OP
height _ PN
base ON
Three-figure tables of the values of these ratios are given
in the next article for angles between o and 90 , at intervals
of one-tenth of a degree.
20
PRACTICAL PLANE GEOMETRY
CHAP
14. Trigonometrical Tables.
SINES OF ANGLES
0"
10
20
30"
40
50
60
70
80
1 2' 3
4 5 6
T 8 9"
Difference to be added.
1 -2 -3
4 -5 -6
7 -8 -9
.ooo
.174
342
.500
643
.766
.866
94o
.985
.017 .035 .052
.191 .208 .225
358 -375 -39 1
.515 .530 .545
.656 .669 .682
777 -788 -799
.875 .883 .891
.946 .951 .956
.988 .990 .993
.070 .087 .105
.242 .259 .276
407 -423 -438
559 -574 -588
.695 .707 .719
.809 .819 .829
.899 .906 .913
.961 .966 .970
995 -996 -99 8
.122.139 -156
.292 309 .326
.454.469 .485
.602.616 .629
73W43 -755
.839.848 .857
.920.927 .934
.974.978 .982
.999 .999 1.000
235
235
235
1 3 4
1 2 4
123
112
Oil
000
7 910
7 8 10
689
679
5 6 7
4 5 6
3 4 4
223
111
12 14 16
12 13 15
11 12 14
10 11 13
9 10 11
7 8 9
567
3 4 4
111
COSINES OF ANGLES
1 2 3
4 5 6
7" 8" 9
Difference to be subtracted. \
V -2 -3
4" -5 6
V -8 9
10
20
30
40
50
60
70
80
1.000
985
.940
.866
.766
.643
.500
342
174
1.000 .999 -999
.982 .978 .974
934 -927 -920
.857 .848 .839
755 -743 -731
.629 .616 .602
485 -4 6 9 -454
.326 .309 .292
.156 .139 .122
998 -996 -995
.970 .966 .961
.913 .906 .899
.829 .819 .809
.719 .707 .695
.588.574 .559
.438 .423 .407
.276 .259 .242
.105 .087 .070
-993 -99o .988
.956 .951 .946
.891 .883 .875
799 -788 .777
.682 .669 .656
545-530.5I5
39i -375 -358
.225 .208 .191
.052 .035 .017
000
Oil
112
I 2 3
I 2 4
1 3 4
2 3 5
2 3 5
2 3 5
1 1 1
223
3 4 4
4 5 6
567
679
689
7 8 10
7 9 10
111
3 4 4
567
7 8 9
9 10 11
10 11 13
11 12 14
12 13 15
12 14 16
TANGENTS OF ANGLES
0-
10
2D
0"
1 2 3
4 5 6
7
8
9
1
Difference to be added. |
1 -2 -3
4 -5 6
7 -8 -9
.COD
.176
34
.017 .035 .052
.194 .213 .231
384 -44 -424
.070 .087 .105
.249 .268 .287
.445 .466 .488
123
306
.510
.141
325
532
.158
344
554
245
246
246
7 910
7 911
811 13
12 14 16 j
13 15 17
15 17 19
30
40
50
577
839
i.ig
.601 .625 .649
.869 .900 .933
1.23 1.28 1.33
.675 .700 .727
.966 1. coo 1.036
I. 38 1.43 1.48
754
1.072
i-54
.781
1. in
1.60
.810
1.150
1.66
3 5 8
4 7 11
112
10 13 16
14 18 21
233
18 21 23
25 28 32
4 4 5
60
i-73
1.80 1.88 1.96
2.05 2.14 2.25
2.36
2.48
2.61
V0 J
80
2-75
5-67
2.90 3.08 3.27
6.31 7.12 8.14
3-49 3-73 4-oi
9.51 11.43 !4-30
4-33
19.08
4.70
28.64
5-i4
57-29
'
GENERAL INTRODUCTION
21
CHORDS OF ANGLES
1
.ooo 1
10
20
174
347
30
.518
40
50
.684
845
60
1. 000
70
80
1.147
1.286
Difference to be added.
go 1 r
9 !-l -2 -3 -4 -5 -6' -T -8 9
.017 .035 .052 .070 .087 .105
.191 .209 .226 .243 .261 .278
.364 .382 .399 J .416 .433 .450
534 -551 -568 j .585 .601 .618
.700 .717 .733 j .749 .765 .781
.8C1 .867 .892 : .908 .923 .939
.015 1.030 1.045 I1.060 1.075 1.089
.161 1. 175 1. 190 I1.203 1. 218 1. 231
.299 1. 312 1.325 11.338 1. 351 1.364
.122
.296
.467
.635
797
954
.140
313
.484
.651
.813
.970
'57
33
.501
.667
.829
9S5
1. 104 1. no 1. 133
1.245 1.259 1.272
1.377 1.389 1.402
9 10
9 10
9 10
8 10
8 10
8 9
7 9
12 14 ID
12 14 16
I2I4I5
12 13 15
II 13 14
II 12 14
IO 12 13
IO II 12
9 IO 12
SINES, TANGENTS, CHORDS, AND RADIANS OF SMALL ANGLES
1
2
3
4
5
fi
7=
.go
9
0"
Diff. to be
added.
6'
12'
18'
24'
30'
36'
42'
48'
54'
.0000
.0017
C035
.0052
.0070
.0087
.0105
.0122
.OI40
oi57
1' .0003
1
0175
.0192
.0209
.0227
.0244
.0262
.0279
.0297
.0314
0332
2' .0006
2
0349
.0367
.0384
.0401
.0419
.0436
.0454
.0471
.O489
.0506
3' .0009
3"
.052
054
.056
.058
059
.061
.063
.065
.066
.068
4' .0012
4
.070
.072
73
75
.077
.079
.080
.082
.084
.085
5' .0015
Examples. 1. Required the sine of 34 5*.
table of sines, we find
Sine of 34 . .559
Add for .5 diff. 7
For the difference 1 1 add
The required angle is
3. Required the cosine of 414 J
Referring to the
Sine of 34.5 . .566
2. Required the angle whose chord is -824,
From the table, the chord of 48 is "813
48.7-
Cosine of 41 . . -755
For excess .4 subtract diff. 5
Cosine of 4 1. 4 . . .7 50
22 PRACTICAL PLANE GEOMETRY chap.
15. Problem. To set off any given angle, say 37 6.
(a) By the protractor. If a circular protractor be used,
then to ensure great accuracy, mark off two points for 3 7 "6,
at opposite ends of a diameter.
(b) By referetice to a table of sines. From the table the
sine of 37.2 is seen to be .61 1.
Now construct the right-angled triangle JVBO, making
JVP=.6n,PO=i. Thus set off NP= 6. 1 1 andi 5 (9=io
on the \" or |" scale.
The angle PON equals 37. 6.
(c) By reference to a table of cosines. From the table we
find cos 37. 6 to be .793.
To any convenient scale set off ON= .793 (or 7.93),
and make OP=\ (or 10).
The angle PONwW be 3 7. 6.
(d) By reference to a table of tangents. We find from
the table that tan 3 7. 6 = .7 70.
Mark off ON= 1 (or 10), and JVP=.tjo (or 7.70);
join OP.
Then the angle PON= 37. 6.
(e) By reference to a table of chords. The chord of 3 7. 6
is found to be .645.
With centre O, radius unity (or 10), describe the
arc PP.
With centre P, radius .645 (or 6.45), cut this arc in P ;
join OP.
Then angle POJV= 3 t.6.
(/) By means of a scale of chords. A scale of chords
marked CHO is generally given on a rectangular protractor.
With centre 0, radius o to 60, describe the arc PP.
With centre P, radius o to 37.6, cut this arc in P.
Join OP, then the angle POJV= 37. 6.
16. Problem. An angle AOB being given, to measure
it in degrees.
With centre 0, radius unity (or 10 read as 1) on any
convenient scale, describe the arc PP. Draw the chord
PE and the perpendiculars PJV, EQ.
GENERAL INTRODUCTION
23
16
(a) Measure the angle directly with the protractor
(b) or measure PN and refer to the table of sines ;
if) or measure ON and refer to the table of cosines ;
if) or measure QE and refer to the table of tangents ;
(e) or measure PE and refer to the table of chords ;
if) or measure PE on the scale of chords, the arc PE
having been struck with the radius o to 60.
Examples. 1. Draw an isosceles triangle with sides of 5", 5",
and 2.2". Determine and measure the vertical angle by each
of the six methods of this article. Take the mean of these,
and observe by how much each separate result differs from the
mean. A/is. The angle is 25. 4 .
2. The sine of an angle is .820, what: is the tangent? Aris. 1.44.
Note. This example should be worked in two ways : first, by
construction ; secondly, by an inspection of the tables.
24 PRACTICAL PLANE GEOMETRY chap,
17. Miscellaneous Examples.
The student, from his previous study, should be able to work such
examples as the following :
1. Determine by construction a perpendicular through a given
point to a given line, for various positions of the point and
restrictions as to the length of the line.
2. Through a given point draw a line to meet a given line at a
given angle.
3. Draw a triangle, having given {a) the three sides ; (/') two sides
and the included angle ; (c) two sides and an angle opposite to
one ; (</) two angles and the intermediate side ; (<") two angles
and a side opposite to one of them.
4. Construct an irregular quadrilateral, having given (a) the four
sides and one diagonal ; (/') the four sides and one angle ; (r)
three sides and the two diagonals ; (J) three sides, one diagonal
and one angle ; (e) three sides and two angles, etc., etc.
5. Construct an irregular pentagon, having given the five sides and
two diagonals ; the five sides and two angles, etc.
6. Construct any irregular polygon, having given adequate data as
to the sides, diagonals, and angles.
7. ABCDE are the five consecutive corners of an irregular pentagon.
Construct the figure, having given
Sides- .// = 2.qi"; BC=2.Sl"; C/^3.56"; ^^=1.34".
Diagonals AC = 4. 63"; ^ = 3.98". Angle DEA = 133.2.
The polygon is not to have any internal angles greater than
180 (called re-entrant angles).
Measure the remaining side and diagonals.
Am. EA = 3.04"; CE = 4. 57; J) A = 3. 66".
8. Set out all the pentagons which comply with the data of Ex. 7,
re-entrant angles being allowed.
Note. There are four different polygons with re-entrant angles.
9. Copy a given triangle so that a specified side shall have an
assigned position.
10. Copy a given quadrilateral so that a specified side or diagonal
shall occupy an assigned position.
11. Copy any given rectilinear figure so that a specified line shall
have a given position on the paper.
12. Draw a triangle similar to a given triangle, the length and
position of one side of the former being given.
13. Copy the figures on page 9, (a) the same size ; {!>) double size.
14. Copy the figures on page 27, (a) the same size ; (/>) double size.
15. Reduce any given quadrilateral to a triangle of equal area ; and
reduce the triangle to an equivalent rectangle.
16. Find the centre of any given circle or circular arc.
GENERAL INTRODUCTION 25
17. Describe a circle 4^" diameter. Suppose this circle to be
given to you on the paper, and that you have only a pencil
and a ruler with two parallel edges 1 j" apart ; show how the
centre of the circle can be obtained. (1888)
18. Determine a circle to pass through three given points, or draw
the circle which circumscribes a given triangle.
19. Draw a circle to touch three given lines, or determine the
inscribed and escribed circles for a given triangle.
20. Draw a tangent to a given circle from a given point on the
circumference.
21. Find the point of contact of a tangent to a circle drawn from
an external point.
22. Draw a circle of given radius to touch (a) a given line at a
given point ; (6) a given circle at a given point.
23. Draw a circle of given radius {a) to pass through two given
points ; (6) to touch two given lines ; (c) to pass through a
given point and to touch a given line.
24. Draw a circle of given radius (a) to touch two given circles ;
(6) to touch a given line and circle ; (c) to pass through a given
point and touch a given circle.
25. 1 >raw a circle of given radius which shall have its centre on a
given line, and touch (a) a second given line ; (/') a circle.
26. Illustrate the theorem of Note 1, Prob. 21, in the following way :
Draw on tracing-paper two lines AP, AQ meeting at any
angle, say make A = 30, and AP, AQ, 3" and 4". Between
AP and AQ draw a series of lines P t Q v 7>. 2 Q.,, P 3 Q 3 , . . .
parallel to J'Q. These may be about J" apart" Glue a small
piece of paper on the tracing at A for strength.
Next draw in ink on paper any figure, say a semicircle 3"
diameter. Mark a point O inside the semicircle, say near the
middle ; place the tracing with A at 0, and insert a pin at
this point.
Then rotate the tracing, and as points on one of the lines,
say on AP, come in succession on the boundary of the figure,
prick through the corresponding points on the other line AQ.
The locus of the latter will be a semicircle, larger in the linear
ratio 3 : 4, and turned through an angle A about O.
Again pin the vertex A of the tracing to a point outside the
semicircle, say near one end. Rotate, and as the points Q
come on the boundary, prick through the corresponding points
P. A copy of the semicircle will again be obtained ; this time
reduced in the ratio 4 : 3, and turned through an angle A.
CHAPTER II
SIMILAR RECTILINEAL FIGURES AREAS
18. Similar polygons. Definition. Two equiangular
polygons which have the sides about their equal angles
proportional are said to be similar.
Thus the pentagons ABCDE and abcde are similar, being equi-
angular and having
AB :BC :CD : DE :EA=ab: be : cd : de : ea ;
AB _BC _CD _DE EA
ab be cd de ea '
Except in the case of the triangle, polygons may be equiangular
without being similar, for example, ABCDE and A'B'C'D'E'. And
evidently, with the same reservation, the sides may be proportional
without the corresponding angles being equal.
We now give some of the more important properties of similar
figures, on which the constructions of many problems are based.
An instrument called a pantograph, for the tracing of similar figures,
is sometimes used.
Theorem i. If two lines be cut by a series of parallel lines,
the ratios of corresponding segments are all equal.
_ KL LM MN KN LN OK OL
thus ,,=, = = . =f = = = . . . etc.
// I in mil kn hi ok ol
Theorem 2. If two polygons have their sides respectively
parallel, and the lines joining their corresponding corners all
converging to a point, the figures are similar ; and conversely.
ch. ii SIMILAR RECTILINEAL FIGURES AREAS
w
In
\
m\
im
L \
I
Thus the polygons PQRS and pqrs which satisfy this condition are
similar, and the following relations hold :
PQ_ Q_#s_sp_qp_ OQ _ OR _ OS
pq qr rs sp Op Oq~ 0r~~ Os'
The point O is called the pole.
The student will find many illustrations of this theorem. Thus in
the figure of Art. 6 the pole is situated within the polygon. In
Prob. 36 the pole coincides with one corner A. In Prob. 24 the
pole is inaccessible ; it might be at an infinite distance away, when the
lines through it would be parallel.
Theorem 3. The areas of similar figures are propor-
tional to the squares on any two corresponding sides.
Thus
area PQRS
area pqrs
{PQf
KQK)
{qr?
{OPf
(Op?'
etc.
or
area ONn _ (JVn) 2 _ {ON?
area OKk ~JM?~{ORy i= CtC "
28 PRACTICAL PLANE GEOMETRY chap.
19. Problem. Two triangles ABC, DEF are given.
It is required to draw a triangle def with its vertices
d, e, f in BC, CA, AB, and its sides de, ef, fd parallel to
DE, EF, FD.
Between AB and AC draw any line F ' E' parallel to
FE ; draw FD, E ' F>' parallel to FD, ED.
Join AD', and let this line (produced) meet BC in d.
Through d draw de, ^/"parallel to DE, EF, and join ef.
Then def is the triangle required.
20. Problem. A triangle ABC and a quadrilateral
DEFG are given. It is required to draw a quadrilateral
defg similar to DEFG, with its side de in AB, and its
vertices f, g in BC, CA respectively.
Copy the figure DEFG in the position D'E'FG' ; that
is, with D'E' in AB, and G' in AC.
Draw AF' to intersect BC inf.
Then draw/?-, gd, fe parallel to F G' , G'D', FE'.
We thus obtain the inscribed quadrilateral as required.
21. Problem. Two triangles ABC, DEF are given.
It is required to draw a triangle def similar to DEF, with
f at a given point in AB, and d, e in BC, CA respectively.
From the given point / draw fE' to meet BC at an
angle fE B FED. From E' draw E'e, making the angle
CE'e = DFE.
Join ef. From e draw ed, where the angle fed = FED.
Join df. Then def is the triangle required.
Note I. This solution is based on the following theorem.
Theorem. If a triangle of varying size but constant shape rotate or
swing about one vertex A, while a second vertex P traces any figure,
then the third vertex Q traces a similar figure turned through an angle
A, the linear dimensions of the two figures being in the ratio AP : AQ.
The figure shows two positions fD'E, fde of such a triangle. As
one vertex moves from D' to d, the other moves along E'e, inclined
at /3 to D'd. See also Ex. 26, p. 25.
Note 2. With modified data some of the points would be situated
in the sides AB, BC, CA produced. The student will now be able to
work examples relating to triangles, squares, parallelograms, etc., in-
scribed in and circumscribed about triangles.
ii SIMILAR RECTILINEAL FIGURES AREAS
29
Examples. 1. Draw a
BC= 3 .S", CA = 2.$'
triangle ABC, making AB = 4",
In ABC inscribe an equilateral
triangle with one side inclined at 45 to AB.
In the triangle of Ex. 1 inscribe a square with one side in AB.
In ABC inscribe an equilateral triangle with one vertex bisecting
AB.
In ABC of Ex. 1 inscribe a similar triangle abc, with c at the
middle point of BC, a in AB, and b in AC.
In ABC inscribe a parallelogram with sides in the ratio 2:3, and
included angle 60, a longer side lying in BC.
Draw a circular arc, 3" radius, and two radii including 6o. In
this sector inscribe a square with a side along a radius.
30 PRACTICAL PLANE GEOMETRY chai>.
22. Problem. Two lines AB and CD and a point P
are given. It is required to draw through P a line
terminated by the given lines, and divided at P into two
segments which are in a given ratio, say 2 : 3.
Through P draw any line meeting one of the given
lines, say AB, at E.
Bisect PE at G, and set off PE= 3 PG. Draw ES
parallel to AB, meeting CD in S.
Then the line through 6" and P meeting AB in R is
the one required.
For, since the triangles PES, PER are similar,
PR _PE _2PG _2
P~S~PF~'^PG~3
Note. If PF were made equal to PE, then SR would be bisected
at P.
23. Problem. To draw the straight line bisecting the
angle between two given straight lines AB and CD, the
intersection of the latter being inaccessible.
Draw any line EF between the given lines, and bisect
the angles AEE and CFE by lines EO and FO meeting
at O.
Also bisect the angles BEE and DEE by lines EO'
and EO' meeting at O'.
Then the line 00' will be the line required.
For it is obvious that O is equidistant from AB, EF, and CD ;
so also is the point O'.
Hence 00' bisects the angle between AB and CD.
24. Problem. Having given two straight lines AB
and CD and a point P, it is required to draw the straight
line through P, such that the three lines converge to the
same point, the point being inaccessible.
Take any convenient points E and E in AB and CD
respectively, and join EF, FP, and PE.
Parallel to EF draw any corresponding line GH; draw
GQ, HQ parallel to EP, FP. Join PQ.
Then the lines AB, CD, and PQ if produced would all
meet at the same point.
II
SIMILAR RECTILINEAL FIGURES AREAS
P\
Q
H B
Examples. 1. Draw a quadrilateral ABCD of the following
dimensions :
Sides AB=i.f; BC=t,.o"; CD =3.40"; ^ = 3.30".
Diagonal BD = ^.^.". Draw the two bisectors of the angles
between AB, DC, and AD, BC.
2. Let the diagonals of ABCD, Ex. I, intersect in P. Through
P draw the two lines which converge respectively to the same
points as the pairs of opposite sides.
3. Through the point P, Ex. 2, draw the two lines between the
pairs of opposite sides of the quadrilateral which are bisected
at P.
4. Draw an equilateral triangle ABC of 3" side, and take points D,
E in AB, AC distant 1.9" and 1.8" from A. Determine the line
through A which converges to the same point as BC and DE.
32 PRACTICAL PLANE GEOMETRY chap.
25. Problem. To find a fourth proportional D, to three
given lines A, B, and C ; that is, to find a line D such
that A : B : : C : D.
Take any two lines intersecting at a point O ; along one
set off OA, OC equal to A and C respectively ; and along
the other set off OB equal to B. Join AB, and draw
CD parallel to AB.
Then OD is the required length of D.
If AE be drawn parallel to CB, then OE is the length
of a line E such that C : B : : A : E.
26. Problem. To find a third proportional C, to two
given lines A and B ; that is, to find a line C such that
A : B : : B : C.
Take any two lines intersecting at O ; along one set off
OA, OB' equal to A and B respectively ; and along the
other set off OB equal to B.
Join AB and draw B'C parallel to AB. Then OC is
the required length of C, such that A : B : : B : C.
27. Problem. To find a mean proportional C between
two given lines A and B ; that is, to find a line C such
that A : C : : C : B.
Take a point O in any line, and set off in opposite
directions OA equal to A, and OB equal to B.
Describe a semicircle on AB. Then OC, drawn perpen-
dicular to BA, is the required length of C (Euc. VI. 13).
C is also called the geometrical mean between A and B.
28. Problem. To divide a line AB in extreme and
mean ratio ; that is, to find a point C in AB such that
AB : AC : : AC : CB.
Bisect AB at D ; draw BE perpendicular to AB, and
equal to BD, and join EA. With centre E describe arc
BE; and with centre A describe arc EC.
This gives C, the required point of division (Euc.
VI. 30).
II
SIMILAR RECTILINEAL FIGURES AREAS
33
By
A i
c /
A\
V
E B
D
A / \
\ \
26
P
Q
R
25
B C
A
B V
B
-Bi
n
s
s
%
/
\
/
/
\
1
\
1
\
I
t
\
1
I
J
27
AT
\ \
-< i-
A
E
D C 28 ^
Examples. 1. Find the fourth proportionals to P, Q, R ; to
A, Q, A; and to Q, P, R. Ans. .92"; 2.7S" ; 1.45".
2. Find the third proportionals to P and Q ; to Q and A; and to
A' and P. Ans. 1.28"; 2.53"; 3.48".
3. Find the mean proportionals between P and (2 ; Q and A' ; and
A and P. Ans. 1.79"; 1.36"; 1.52".
4. Divide A in extreme and mean ratio, and give the value of the
ratio. Ans. 1.62.
D
34 PRACTICAL PLANE GEOMETRY chap.
29. Problem. To find the harmonic mean between
two given lengths a and b ; that is, to determine a length
k such that b-k:k-a::b:a.
From any point H in a straight line mark off ffA, HB
equal to a and b.
Through A and B draw any two lines AO, BO per-
pendicular to one another. Join OH and set off the angle
A K equal to the angle A Off.
Then HK is the required length of k, the harmonic
mean between a and b ; and ffA, HK, HB are in
harmonic progression.
.Votes. The relation b 'k ~. k a = b : a is equivalent to
KB _HB _ AH _BH .
AK~ HA~' ,OX J^A~ ~BK '
and comparing these equations we see that BA', BA, BH are also in
harmonic progression.
The line AB is said to be harmonically divided in H and A~ and
the line HK in A and B. And four such points are spoken of as a
harmonic range.
The series of lines or pencil of rays formed by joining II, A, K, B
with any point 0' is called a harmonic pencil, because any transverse
line or transversal can be shown to be cut harmonically by the pencil.
Thus H ', A', A"', B" and B x , II 1 , ./ x , A\ are harmonic ranges.
Stated formally
Theorem I. A harmonic pencil divides all transversals har-
monically.
The alternate points or rays are said to be conjugate to one another.
Thus A and B, or 0.1 and OB are conjugate. Likewise OH and 0A~.
Theorem 2. A transversal which is parallel to a ray is bisected
by the conjugate.
Thus the transversal A'^H^ is parallel to O'B, and is bisected in
A . We also have
Theorem 3. If a transversal parallel to any one of the four rays
of a pencil is bisected, the pencil is harmonic.
Thus if OC bisect any angle A OB, and OD be taken perpendicular
to OC, the pencil is harmonic because any transversal parallel to OD
or OC is bisected.
A cotnplete quadrilateral affords an example illustrating harmonic
pencils. See the figure.
Let the diagonals of any quadrilateral PQRS intersect in N, and
the pairs of opposite sides produced in Z, M. Join 71, J/, N. Then
the pencils radiating from L, M, N are all harmonic.
II
SIMILAR RECTILINEAL FIGURES AREAS
35
Examples. 1. Take a and b each double the length given in the
figure above, and find the harmonic mean k.
2. Find the arithmetical mean r, and the geometrical mean e;
between a and b, and vertify the theorem that /-, g, and / are in
geometrical progression.
3. Draw any complete quadrilateral PQRSLMlV, and test whether
the pencils through L, M, A^are harmonic, by applying Theorem
3 of this article.
4. Taking the lines a and b in the figure above, find a line c such
that a, b, c are in harmonic progression. Ans. - 2.46".
5. Draw OA, OH, OJ3, making the consecutive angles A OH,
HOB equal to 25 and 55 respectively. Find the fourth ray
of the harmonic pencil, and measure the angle it makes with
OB. Ans. 70. 8.
6. Through the point b, Fig. 6, p. 47, draw a straight line cutting oa,
oc in d and e, such that the triangles odb, obe are equal in area.
7. Draw a line AB 2" long, and divide it internally and externally
at K and H into segments which have the same ratio, say 3 : 4.
36 PRACTICAL PLANE GEOMETRY chap.
30. Problem. To reduce a given polygon ABCDEF
to a triangle of equal area.
Join D to B, A, and F Draw CG parallel to DB to
meet AB produced in G, and join DG. Draw EH parallel
to DF to meet AF produced in H, and draw HK parallel
to DA to meet BA produced in K. Join DK.
Then DKG is a triangle having an area equal to the
polygon.
This solution is based on the theorem that triangles on the same
base and between the same parallels are equal in area.
31. Problem. To construct a square equal in area to
a given rectangle.
Determine a mean proportional between the sides of
the rectangle (Prob. 27); this is equal to the side of the
required square.
32. Problem. To construct a square equal in area to
a given triangle.
Let a rectangle be drawn, one side of which is equal to
the base of the given triangle, and an adjacent side equal
to half the altitude.
The rectangle so drawn is equal in area to the triangle.
Hence the problem reduces to the last one.
33. Problem. To construct a square equal in area to
a given polygon.
Reduce the given polygon to an equivalent triangle
(Prob. 30). Then apply Problem 32.
34. Problem. To construct a rectangle equal in area
to a given rectangle, and having one side equal to a
given line.
Let ABCD be the given rectangle, and AF the given
side of the required rectangle.
Draw FF equal and parallel to BC ; join FA cutting
BC at H; through //draw MK parallel to AF.
Then ME is the required rectangle.
II
SIMILAR RECTILINEAL FIGURES AREAS
37
L -"
,,'-''
//
34 B
35. Problem. To divide a given polygon ABODE into
a number of equal areas by lines drawn from a given point
P in one side. Say into seven parts.
As in Prob. 30, draw EF parallel to PA ; and DG,
GH parallel to PC, PB, thus reducing the polygon to an
equivalent triangle with vertex P and base FH.
Divide FH into the required number of equal parts.
Then, reversing the construction, draw n' parallel to
FE ; 44', 5 5', 66' parallel to HG ; and 6' 6" parallel to GD.
Join P to i', 2, 3, 4', 5', 6", and the problem is solved.
Examples. 1. Draw a pentagon ABCDE as follows :
Sides AB=i$'; BC=2" ; CD = 2\" ; DE=i\".
Angles ABC=\2o; B CD = 8o ; CDE= 125."
Reduce the figure to an equal triangle with vertex D and
base along AB. Then reduce the triangle to an equal square.
2. Divide the pentagon of Ex. 1 into eight equal parts by lines
drawn through the middle point of DC.
38 PRACTICAL PLANE GEOMETRY chap.
36. Problem. It is required to draw a polygon
similar to a given polygon, and having an area which
bears to the area of the latter a given ratio, say 3 : 5.
Let ABCDE be the given polygon. In any side BA,
produced in either direction, say beyond A, set off Ap and
AP equal to 3 and 5 on any suitable scale.
On Bp, BP draw semicircles as shown ; and from A draw
a line perpendicular to AB, meeting these semicircles in q
and Q. Join QB, and draw qb parallel to QB.
Then Ab is one side of the required figure, and the
latter AbcdeA is completed by drawing the diagonals AC,
AD, and then the sides be, cd, de, parallel to BC, CD, DE.
For (Aqf = AB.Ap, (Euc. III. 35, or Prob. 27),
and (AQ? = AB.AP;
area AbcdeA {Abf {Aqf AB-Ap _Ap _3
then
area ABCDEA {AB) 2 {AQf AB-AP AP 5
37. Problem. To divide a given triangle ABC into
two equal areas, by a line drawn through any given
point P.
Bisect the sides of the triangle in D, E, F.
Join AD, BE, CF and bisect these lines in G, PP, K.
Join HK, intersecting OD in L, and bisect OL in AP.
And through H, M, K draw a fair curve touching BH and
CK2X //and K.
In like manner draw the curves GH, GK.
Through P draw PQ so as to touch one or other of
the three curves, according to the situation of P.
Then PQ divides the triangle into two equal areas.
A T ote. The true curves should be hyperbolas, to which the sides of
the triangle are asymptotic (see Chapter IV.). The bisection of OL in
M makes them parabolas, from which the true curves for such small
arcs do not visibly differ.
Alternative Solution. Along AB, AC set off AN, AN each equal
to the geometrical mean between AB and AE. Bisect NN in T. Join
A T and produce to S, making AS=AN Then T is the vertex, S a
focus, and A the centre of the hyperbola KH, and the tangent through
P may be found by a construction analogous to that of Prob. 94.
SIMILAR RECTILINEAL FIGURES AREAS
39
Examples. 1. Draw a triangle ABC having given AB^.o",
BC=t,.2", CA= 1.8". Divide the triangle into two equal
areas by a line parallel to AB.
2. Divide the triangle of Ex. i into five equal areas by lines parallel
to BC.
Note. Divide AB into five equal parts, and from the points of
division draw perpendiculars to meet a semicircle on AB in
I, 2, 3, 4. With centre A draw arcs through I, 2, 3, 4 to
meet AB in 1', 2', 3', 4'. Draw lines through the latter points
parallel to BC.
3. Draw a pentagon ABCDE as follows :
Sides AB= 1. 7" ; BC=i.3$" ; CD= 1.45"; >= 1.65" ;
EA=.$". Diagonals AC= 2.5", AD=2.2".
Draw two similar pentagons, one 2 and the other 4- the area
of ABCDE.
4. Divide the pentagon of Ex. 3 into four equal areas by lines
parallel to the sides.
5. Divide the triangle of Ex. 1 into two parts of equal area, by a line
passing through a point /"outside the triangle, the position of
/being given by AP= 1" ; BP= 2.8".
4 o PRACTICAL PLANE GEOMETRY chap.
38. Problem. To find the number of square units of
area in any given polygon.
Let ABCD be the square equal in area to the polygon,
obtained as in Prob. 2>e>-
Produce one side DA to *S", making AS one unit of
length. Join SB, and draw BT perpendicular to BS.
Then AT, measured on the scale having AS as unity,
gives the number of square units of area required.
Thus if AS be set off i", the area in square inches is obtained by
measuring A T on the i" scale. Or if AS be I centimetre long, then
AT measured on the centimetre scale gives the area in square centi-
metres.
Note that A T is a third proportional to AS and AB ; and that AB
is a mean proportional between AS and AT.
To construct a square of given area, say 3 square inches, draw a
rectangle 3" long and 1" broad, and reduce to an equivalent square.
Or set off AT= 3", AS= 1", and find the mean proportional AB.
39. Problem. To construct a polygon similar to a
given polygon ABCD, and having an area equal to that
of a given polygon Q.
As in Prob. 40, determine AF, the side of a square equal
in area to Q; and also AG, the side of a square equal in
area to ABCD. (Construction not here shown.)
Set off AB and AG along a line AE, making any angle
with AB ; join GB, and draw Fb parallel to GB.
Join AC, and draw be and cd respectively parallel to BC
and CD. Then Abed is the required rectilinear figure.
area of Abed Al> 2 AF 2 area of Q
For
area, of ABCD AB 2 AG 2 area of A BCD'
It is important that the reader should observe that this problem is
the general case of a distinct type or class of problems, and that the
method of solution is of a general character. Two figures, which may
vary in size and shape, are given, and a third figure has to be deter-
mined, such that it is similar to one of the given figures and equal
in area to the other. This is the type of problem. The data of
the problem may be varied considerably, in so far that two triangles,
rectangles, squares, etc., or any combination of these, may be substi-
tuted for the pair of figures given above ; but the problem remains
cf the same character, and the same method of solution is employed.
ii SIMILAR RECTILINEAL FIGURES AREAS 41
D 39
Such observations should render it unnecessary to explain in detail
the solutions of further problems of the same kind, such as the
examples below.
It may here be remarked that the student should acquire the
habit of closely observing the type of any problem which comes under
his consideration, and constantly aim at classifying problems, with the
object of detecting general methods where such may exist, rather than
to regard each problem as having no relation or resemblance to
previous problems. In this manner the power of successfully attacking
examples may often be greatly increased.
Examples. 1. Construct an isosceles right-angled triangle equal
in area to a given regular pentagon 1 j" side.
2. Construct an isosceles triangle, vertical angle 40 , equal in area
to a triangle whose sides are 2^", i||", and 2. 1".
3. Construct a rhombus, one angle of which is 6o, equal in area
to a given rectilineal figure ABCD, of which AD is 2", angle
BAD 105 , AB 2", BC 3I", DC 1.7".
4. Determine an equilateral triangle having an area of 4 square
inches. Measure the side. Ans. 3.04".
5. Construct a regular heptagon having an area of 5 square inches.
6. Construct a rectangle equal in area to a hexagon of It" side,
the ratio of the sides of the rectangle being 2 : 3.
7. Determine graphically the area of the quadrilateral ABCD of
Ex. 3, in square inches and also in square centimetres. Ans.
4.62 ; 29.8.
42 PRACTICAL PLANE GEOMETRY chap.
40. Problem. Having given two similar rectilinear
figures, it is required to construct a similar figure, the
area of which shall be equal to the sum or difference of
the areas of the given figures.
Let ABCDE, Abcde be the two given similar figures.
For the sum draw BN at right angles to AB, and
make BN= Ab.
With A as centre, describe the arc NB' to meet AB
produced in B' ; draw B'C, CD', D'E' respectively parallel
to BC, CD, and DE, to meet AC, AD, AE produced in
C ' , D ' , and E ' . Then A'B'CD'E' is the required figure.
For the difference describe a semicircle on AB as
diameter.
With centre B, radius Ab, cut this semicircle in M;
With centre A, radius AM, draw the arc MB".
Then a polygon with AB" as one side, and similar to
the others, is the figure required.
Note. These constructions are based on the following theorem, of
which Euc. I. 47 is a particular case.
Theorem. If similar polygons be described on the three sides of a
right-angled triangle ABC, so that AB, BC, CA are corresponding
sides of the three figures respectively, then the areas of the polygons
described on the sides AB and BC are together equal to the area of that
described on the hypothe?iuse AC.
41. Problem. Having given any three rectilinear
figures, which may be denoted by A, B, and C, to deter-
mine a fourth figure which shall be similar to C, and
have an area equal to the sum or difference of the areas of
A and B.
Determine, by means of Prob. 39, a figure E similar to
A, and equal in area to B.
By Prob. 40 determine a figure F similar to A (or E),
and having an area equal to the sum or difference of the
areas of A and E.
Finally, by means of Prob. 39, determine a figure G
similar to C, and having an area equal to that of F; then
G is the required figure.
ii SIMILAR RECTILINEAL FIGURES AREAS 43
G E.
C! A
B
\h
42. Problem. To find a rectangle equal in area to a
given trapezoid ABCD.
Let AD, BC be the parallel sides of the trapezoid.
Bisect AB, CD in E, F. Then EF is parallel to AD
and BC, and is midway between them. Through E and
F draw GH and KL perpendicular to EF.
Then the rectangle GL is equal in area to the trapezoid,
and the area is equal to EFx KL, or - x KL.
2
Examples. 1. Draw two equilateral triangles respectively equal
to the sum and difference of two equilateral triangles of 2" and
3" sides.
2. Draw two circles respectively equal to the sum and difference
of two circles of lA" and 2|" diameters.
3. Draw a square equal to the sum of the areas of two equilateral
triangles of 1.7" and 2.8" sides.
4. Draw an equilateral triangle equal to the difference of two
squares of 1.2" and 2.7" sides.
5. Draw a square equal to the sum of an equilateral triangle and
a regular pentagon, each of ii" side.
6. Draw an equilateral triangle equal to the difference of a square
and a regular hexagon, each of 2" side.
44 PRACTICAL PLANE GEOMETRY chap.
43. Problem. To determine approximately the area
enclosed by a given irregular curve AB, a base line CD,
and two perpendicular ordinates CA, DB.
A perpendicular erected at any point on the base CD of the figure
is called an ordinate.
Draw a series of equidistant ordinates between CA and
DB, dividing the area into a number of strips of equal
width ; in this case six. One of these strips is shown with
shade lines.
Draw a second series of intermediate ordinates, midway
between those first drawn, and let y v y 9 , . . . y 6 denote
their lengths.
Determine the mean ordinate y,
_ Jl+J'-2 + - -+}\
6
6
Take CK equal to y m and through K draw KL parallel
to CD. Then the area of the rectangle KD is equal to
area required.
The result is only approximate, because the upper boundary lines of
the strips are curved instead of straight. By increasing the number of
strips, and so making the width of each less, the error can be reduced
to a very small proportion of the total area.
We now state some other rules, available for calculating approxi-
mately the mean ordinate y m ; that is, the height CK of the equivalent
rectangle.
In employing these we require the ordinates which bound the strips
(not the middle ordinates of the spaces). In the figure there are
seven such ordinates to the six divisions, CA being the first, and DB
the last. These may be denoted by A , /i v h. h % , //.,, // 5 , h 6 .
If there were n equal divisions, or strips, there would be ;/ + I
ordinates
"o> "i> "2> " 1> "
With this notation, the rules in question maybe written as follows :
Ordinary rule. Any number of divisions.
?m = I (i U'O + h n) + h +*, + *,+ . . . + V-l}
Simpson's Ji ist rule. Number of divisions even.
When n = 2. y m = \ (// + 4 /i l + h 2 ).
When n~2, or 4, or 6, etc.
?>* = {b + A n +2 (^ 2 + ^4 + - +*-) + 4(* 1 + A 8 + . -+K-l)}-
ii SIMILAR RECTILINEAL FIGURES AREAS 45
Simpsotis second rule. Number of divisions a multiple of 3.
When n
When =3, or 6, or 9, or 12, etc.
J'm = k ( 7/ + Z h \ + 3 A 2 + * 3 ).
+ K-z) + 3(^1 + * a + ; h + h + - + K-*. + K-i)}
Weddle's rule. Number of divisions a multiple of 6.
When n = 6. y m = fa {A Q + K + A t + h Q + 5 (A + h z + h b ) + h 3 } .
When h = 6, or 12, or 18, etc.
J ' = Tcb^ + / ' 2 + - +*+5(*i+*, + .
+ Vl)+^3 + 7 '6 + - -+K-Ji-
Examples. 1. Draw a quadrant of a circle 3" radius. Divide
one of the radii into six equal parts, and draw the ordinates,
which will be parallel to the other radius. Determine the
mean ordinate by each of the rules given, and compare the
results. Set out the equivalent rectangle. Reduce to an
equivalent square. And find the area of the quadrant in square
inches. Ans. Mean ordinate = 2. 36". Side of square = 2. 66".
Area =7. 07 sq. inches.
2. Plot the curve for which thirteen equidistant ordinates, spaced
i" apart, have the following successive values :
".4.1", 4-o", 4.0", 3-9"; 3-55", 3-16", 2.73", 2.34", 2.05",
1.77", i.6o", 1.45", 1.0".
Obtain the mean ordinate by each of the rules given. Draw
the equivalent rectangle, and determine the area under the
curve in square inches. Ans. Mean ordinate = 2. 76". Area =
16.6 sq. ins.
3, In Ex. 2, Art. 156, find the average school attendance for the
six years 1892-97. Ans. 119,100.
46 PRACTICAL PLANE GEOMETRY chap.
44. Miscellaneous Examples.
Note. The figures are to be copied twice size.
*1. Divide the given triangle ABC into four equal parts by lines
drawn parallel to AB. (1896)
*2. Draw a triangle similar to the given triangle ABC, but with a
perimeter equal to the harmonic mean between the lines LAI
and PQ. (1896)
*3. Divide the given line A into two parts, such that the sum of the
squares on them shall be equal to the square on the given line
B. State what limitations are necessary as to the length of B
in order that the problem may be possible. (1S88)
Hint. The solution of a problem is often suggested by an
algebraical investigation.
Thus let x be one part, and therefore A - x the other. Then
by the conditions x 2 + (A - x) 2 = B 2 . Solving which, we find
X = iJyA Ji^Bf-A*}.
Now V2.5 is the diagonal of a square on the side B. If
this be made the hypothenuse of a right-angled triangle, and
A one of the sides, the other side is / (V ' 2B) 2 - A 2 . The
rest of the construction is simple.
*4. Between the lines ab, cd produced if necessary, place a line
parallel to ef in such a position as to form a four -sided figure
with an area of 2 square inches. (r89i)
Hint. Produce ab, cd to meet in o. Find the area of the
triangle ofe. Call this m square inches. Then find the triangle
oFE similar to ofe, the areas of the two having the ratio 2 + m : m.
*5. Draw a line ab parallel to the base AC of the given triangle
ABC, cutting off segments Aa on AB, Cb on CB, such that
their difference Aa Cb shall be equal to one inch. ( 1S95)
Hint. Aa may be found as the fourth term in the following
proportion
AB-BC :AB:: Aa - Cb : Aa. .
*6. Construct an equilateral triangle which has one angular point at
b on the line ob, and the remaining angular points on oa and oc
respectively. (H. 1886)
7. Draw a triangle ABC with the following dimensions. AB = ^",
BC=2^", AC = 2". Inscribe in the triangle a rhombus, one
side lying in AC, and the adjoining sides inclined to AC at
45- (1894)
8. Given two lengths P= 2. 25", Q=2."jo". Find a line whose
length x is such that P 2 = (x Q) x. Write down the value of x.
(1894)
ii SIMILAR RECTILINEAL FIGURES AREAS 47
4 8 PRACTICAL PLANE GEOMETRY chap.
Hint. P is a mean proportional between (x - Q) and x.
Therefore in Fig. 27, page 33, make OB = P, OC=Q. Draw
CA perpendicular to CB. Then AB = x.
9. In Ex. 8 find x if P 2 = (Q-x)x.
*10. Reduce the given hatched figure to a triangle. (1883)
Hint. Reduce the whole figure and the hole to triangles.
Then determine a triangle equal to the difference of the two.
*11. Through the given point p draw two lines cutting the two
given lines in such a way that the included area is i\ square
inches. (!887)
Hint. Through /draw any two lines intersecting the given
lines in a, A ; b, B. Reduce AabB to a square, and find its
area in square inches by Prob. 38. The length of AB required
for an area of 3^ square inches may now be found as a fourth
proportional to area AabB, 3j, AB.
*12. In what sense can areas be represented by straight lines ? If
the area of the given triangle ABC is represented by a line 1"
long, draw a line representing the area DEFG. (1S85)
*13. Reduce the given figure to a rectangle on the base ab. (1886)
Hint. Apply a construction similar to that of Prob. 30. Thus
drawy?, Im, mn respectively parallel to ge, gd, gc, to meet the
lines ed, dc, ca. Join gn, and through its middle point draw a
line parallel to ab. We thus obtain the required rectangle.
*14. Draw a straight line from A and terminating on FG, such
that the sum of the areas included between it and the given
zigzag line on the one side is equal to the sum of those on the
other. (1885)
15. Draw a quadrilateral figure having an area of 3 square inches,
and made up of two isosceles triangles having a common base
which is a mean proportional between the sides of the triangles.
The vertical angle of the smaller triangle is 90 . (1881)
16. A line 1-5" long represents a square of 2" side. Determine a
length of line which would represent on the same scale a
hexagon of if" side. ( x 883)
17. Draw a right-angled triangle with hypothenuse 2" and one side
if". Prove by construction that the square on the hypothenuse
is equal to the sum of the squares on the sides. ^893)
Hint. One method is to apply the construction of Prob. ^8.
18. Draw an equilateral triangle of 2" side, and circumscribe this by
a right-angled isosceles triangle, so that one of the equal sides
of the latter is bisected.
ir SIMILAR RECTILINEAL FIGURES AREAS 49
Cb/ii/ -tfie figures double size
p
E
CHAPTER III
TRIANGLES, CIRCLES AND LINES IN CONTACT
45. Introduction. The problems in this chapter relate
mainly to the contact of lines and circles, and the con-
struction of triangles from adequate data.
Euclid permits a line to be drawn between two definite
points, but does not allow a common tangent to be drawn
to two given circles without having previously determined
the points of contact by a special construction.
This construction may be necessary in a strict system of
deductive reasoning like Euclidean geometry, but does not
add to the accuracy in a scale drawing, as the student may
easily test for himself.
So in "practical" geometry a tangent may be drawn to
a circle from an external point, or a common tangent to
two circles, by simply adjusting the straight-edge and draw-
ing the line without any previous construction. Then if
the point of contact be required, it is necessary to draw the
perpendicular radius.
If a tangent is required at a point on the circumference,
it must be drawn perpendicular to the radius to the point.
With care, a circle may be drawn with a given centre to
touch a given line, without first finding the point of contact.
But in this case it is generally preferable to draw the
perpendicular from the point to the line before drawing the
circle.
ch. in TRIANGLES, CIRCLES AND LINES IN CONTACT 51
Before working the problems of this chapter, the student
should illustrate the truth of the following theorems, by
making accurate drawings to scale, measuring the results
and making calculations where necessary. For several
reasons this is a most valuable form of exercise ; by com-
paring his results with the true ones he may observe how
small need the errors be which he introduces into his
graphical work. It will not be necessary to add that he
must see that his pencil is in proper condition.
Theorem 1. Angles in the same segment oj a circle are
equal to each other (Euc. III. 21).
Theorem 2. The two tangents drawn from a point to a
circle are equal to each other (Euc. III. 17).
Theorem 3. If through any point in the circumference of
a circle a chord and tangent be drawn, the angles between
them arc equal to the angles in the alternate segments of the
circle (Euc. III. 32).
Theorem 4. If any point P be taken inside or outside a
circle and two lines be drawn through P, one cutting the circle
at A and B, and the other at C and T>, the product of PA
and PB is equal to the product of PC and PD (Euc. III. 35).
Also if Pf>e outside the circle and a line PT be drawn to
touch the circle at T, the squa?'e of PT is equal to the pro-
duct of PC and PD, or of PA and PB (Euc. III. 36).
Theorem 5. If a line which bisects the vertical angle A
of any triangle ABC, cut the base BC in D, the ratio of BD
to DC is the same as the ratio of BA to AC (Euc. VI. 3).
Also if a line bisecting the exterior angle at A cut the
base BC produced in D\ the ratio of D'C to D'B is the
same as that of AC to AB (Euc. VI. 3).
D and D' are said to divide BC internally and externally
in the ratio of the sides of the triangle. (Or AB, AD, AC,
AD' form a harmonic pencil. See Prob. 29.)
Theorem 6. If A and B are two fixed points, and a
poitit P move so that the ratio PA : PB is constant, then the
locus of P is a circle.
5 2 PRACTICAL TLANE GEOMETRY chap.
46. Problem. On a given line AB, to describe a
regular polygon ; say a heptagon.
First Method. Produce AB, and with centre B, radius
BA, describe the semicircle AC.
By trial, divide the semicircle into as many equal parts
as the polygon has sides, in this case seven. Join B to
the second point of division from C.
Then B2 is a second side of the required heptagon.
Draw a circle through A, B, 2. This is the circum-
scribing circle of the required polygon, and the length AB
should step exactly seven times round the circumference.
Second Method. Produce AB, and set off the angle CB2
equal to 360-=- 7, that is to 51. 4 .
Make B2 equal to BA. Then proceed as before.
Note. To inscribe a regular polygon of n sides in a circle, divide the
circumference by trial with the dividers into n equal parts ; or, find one
side by setting off from the centre an angle equal to 360-=- n.
47. Problem. On a given line AB, to construct a
segment of a circle which shall contain an angle equal to
a given angle a.
Draw CD bisecting AB at right angles, and make the
angle DCE equal to a. Draw AO parallel to CE.
Describe a circle with O as centre, radius OA ; then
that segment of the circle, on the side of AB on which D
lies, is the one required.
If the given angle is a right angle, the point O coincides
with C. If the angle is obtuse, O is on DC produced.
48. Problem. From a given circle to cut off a segment
which shall contain an angle equal to a given angle.
Take a point A on the circumference and draw the
tangent AD.
Make the angle DAB equal to the given angle.
Then the segment ACB is the one required.
49. Problem. In a given circle to inscribe a triangle
the sides of which shall be in a given ratio.
Draw any triangle having its sides in the given ratio.
in TRIANGLES, CIRCLES AND LINES IN CONTACT 53
At any point A in the circumference of the given circle
draw the tangent DE. Make the angles DAB, EAC
respectively equal to two of the angles of the triangle,
join BC. Then the triangle ABC is the one required.
50. Problem. To construct a triangle, having given
the base, vertical angle, and altitude.
Let BC be the given base ; describe on it the segment
of a circle containing an angle equal to the given vertical
angle. Draw DE parallel to BC, the distance between
these lines being equal to the given altitude. If A, A' are
the points in which DE intersects the segment, either of
the triangles ABC, A' BC will satisfy the given conditions.
54 PRACTICAL PLANE GEOMETRY chap.
51. Problem. The perimeter of a triangle is 5" ; the
vertical angle is 70 ; and one of the sides is half the hase.
Construct the triangle.
Take any line AB ; on AB draw a segment of a circle
containing an angle of 70 (Prob. 47).
With centre A, radius half AB, describe an arc cutting
the circle in C. Join CA, CB.
Then ABC is a triangle similar to the one required.
Produce AB both ways; make AE equal to AC, and
BD equal to BC.
Divide a line 5" long into segments having the ratio
EA : AB : BD (Prob. 8). Then these segments are
equal to the required sides of the triangle.
52. Problem. To construct a triangle, having given
the base, one of the hase angles, and the perimeter.
Let BC be the given base. Draw BE such that the
angle CBE is equal to the given base angle. Along BE
set off BD equal to the sum of the remaining sides.
Join DC ; draw EA bisecting DC at right angles, and
join AC. Then the triangle ABC is the one required.
53. Problem. To construct a triangle, having given
the base 2", perimeter 5", and area 0-85 square inch.
In the method here given we first determine the side of a square
0.85 square inch in area, and then obtain a rectangle equal in area to
this square, one side of the rectangle being equal to the given base, 2".
The other side of this rectangle will be equal to half the altitude of the
required triangle.
Make OS=i" and <9iV= 0.85". On SN describe a
semicircle, and draw OX perpendicular to SJV; then OX is
the length of the side of the square in question.
Now make OR- given base =2". Bisect RX at right
angles by the line YO\ and with O' as centre and radius
O'X describe the arc XT; then the area of the rectangle
RO.OT= (OX) 2 = 0.85 square inch, and therefore OT=
half altitude of required triangle. The problem may now
be completed as in Prob. 55.
in TRIANGLES, CIRCLES AND LINES IN CONTACT 55
R
51
D,a
x
YJ"
S 0'
53
D
T N
Examples. 1. Describe a regular pentagon on a line if" long.
2. In a circle 2|" diameter inscribe a triangle two angles of which
are 45 and 65 .
3. Construct a triangle having the base if, perimeter 4|", and
area .95 of a square inch.
56 PRACTICAL PLANE GEOMETRY chap.
54. Problem. To construct a triangle, having given
the vertical angle, altitude, and perimeter.
Draw two lines AD, AE containing an angle equal to
the given vertical angle. With centre A, and radius equal
to the given altitude, describe the arc MN,
Along AD and AE set off AF and AG each equal to
half the given perimeter.
Draw FO and GO perpendicular to FA and GA, and
with as centre, describe the arc GF. If now a com-
mon tangent be drawn to this arc and the arc MN, so
as to meet AD and AE in B and C respectively, then
ABC is the required triangle.
55. Problem. To construct a triangle, having given
the base, altitude, and the perimeter.
Let F'Fbe the given base; bisect F'F in C, and make
CA = CA' = half the sum of remaining sides, i.e. = ^ (peri-
meter - base). Draw CE perpendicular to AA', and with
T^as centre and CA as radius, describe an arc intersecting
CE in B. Make CQ = given altitude, and CG = CB.
Join GQ, and draw A' K parallel to GQ. On A A' describe
a semicircle, and draw KP' parallel to A' A. Draw F'F
parallel to BC, and QF parallel to KP' , then PFF' is the
required triangle.
Note. The point P is on an ellipse, major axis A A', foci F, F'.
The above construction is based on the properties of an ellipse.
56. Problem. To construct a triangle, having given
H, K, L, the lengths of the three medians.
A median is a line drawn from any vertex to the middle
point of the opposite side.
Make AD = Ff, and produce it to G such that DG =
DO = IAD.
^'ith O and G as centres, and f K and L as radii
respectively, describe two arcs intersecting at B.
Join BD and produce it to C, making DC ' = DB, then
ABC will be the required triangle.
Note. For explanation, compare the construction with the triangle
abc in which the medians have been drawn.
in TRIANGLES, CIRCLES AND LINES IN CONTACT 57
56
-JG'
58 PRACTICAL PLANE GEOMETRY chap.
57. Problem. To construct a triangle, having given
(a) the hase, vertical angle, and the ratio of the sides,
say 3:5; (h) the base, altitude, and the ratio of the
sides, 3 : 5.
(a) Let AB be the given base. Select any convenient
unit, and construct the triangle AP'B, making BP' = 3
units, and AP' = 5 units. Bisect the angle AP'B by the
line P'D, and draw P' D' perpendicular to P ' D.
On DD' as diameter describe a circle ; this circle must
contain the vertex of the required triangle. On AB
describe the segment ABE of a circle, containing an angle
equal to the given vertical angle (Prob. 47). Then the point
P in which this segment intersects the circle on DD' is
the required vertex, and APB the required triangle.
(b) Proceed as in (a) so far as obtaining the circle on
DD'. Draw AF perpendicular to AB, making AF= the
given altitude ; draw FG parallel to AB, intersecting, in
P, the circle on DD', then APB is the required triangle.
The construction for this problem is based on Theorems 5 and 6,
Art. 45.
58. Problem. To construct a triangle equal in area
to a given triangle ABC, and having two of its sides
respectively equal to two given lines D and E.
Draw AF perpendicular to BC, bisect AF in G, and
complete the rectangle BHKC. Draw LM equal to one
of the given lines, say D, and mark off LN equal to BC.
Draw MO perpendicular to ML, making MO equal to
FG ; join LO, and draw NP parallel to MO, then the
rectangles QRML and BHKC are equal in area (Prob.
34). Produce NP to ,.9, making PS equal to PN, and
draw ST parallel to LM. With L as centre, and the
other given line E as radius, describe an arc intersecting
TS in V, then the triangle VLM satisfies the required
conditions.
Note. The arc described with L as centre would intersect ST
produced in V, hence there are two solutions.
in TRIANGLES, CIRCLES AND LINES IN CONTACT 59
E
A\ 2K IB
D
N M
Examples. 1. Construct a triangle having the vertical angle
55j altitude 2", and perimeter 5 inches.
2. Construct a triangle having the base 2'', altitude \\" , and
perimeter 5 inches.
3. The three medians of a triangle are i|", 2j", and 3" ; draw the
triangle.
4. Construct a triangle having its base 3", vertical angle 120, and
the ratio of sides 3 : 5.
5. Construct a triangle having its base 2", altitude 2j", and ratio
of sides 4 : 7.
6. Draw a triangle with sides of 2", 2\", 2|", and construct another
triangle having the same area, two of its sides being 2" and 3".
60 PRACTICAL PLANE GEOMETRY chap.
59. Problem. To describe a circle to pass through two
given points P and Q, and touch a given line AB.
Join P and Q, and produce PQ to cut AB in P.
Along AB set off R T (either to the right or left) equal to the mean
proportional between PQ and PP.
Draw TD at right angles to AB, and bisect PQ at right angles by
NO, meeting TD in 0.
Then is the centre of one circle satisfying the required conditions.
60. Problem. To describe a circle which shall have
its centre on a given line CD, pass through a given point
P, and touch a given line AB.
Let fall a perpendicular /Wfrom P on to CD ; and produce PN
to Q, making NQ equal PN. Then, by Prob. 59, determine a circle
to pass through P and Q and touch AB.
61. Problem. To describe a circle to pass through a
given point P, and touch two given lines BA, AC.
Draw AD bisecting the angle CAB.
With any point O 1 in AD as centre, describe the circle touching
AC and AB.
Draw AP, and produce it to cut this circle in P' (and P 1 not shown).
Join P'O, and draw PO parallel to P'O.
Then O is the centre of one circle satisfying the conditions.
62. Problem. Between two given lines AB, AC, to
inscribe a succession of circles in contact.
Draw AD to bisect the angle BAC. Take any point O on AD as
centre, and describe a circle touching AB and AC, and cutting AD in
E, F.
Draw OG perpendicular to AB, and join EG, FG. Draw EH,
HP respectively parallel to FG, GO ; and FK, KQ to EG, GO.
With centre P and Q, describe circles through E and F.
The three circles will touch the two lines and each other as required.
The circles may be extended by repeating the construction.
63. Problem. Two straight lines AB and CD are
given, their point of intersection being inaccessible. It
is required to describe a circle which shall touch these
lines and pass through a given point P.
By Prob. 22, draw FF to bisect the angle between AB and CD.
With any centre O in EF, draw the circle which touches AB.
By Prob. 24, draw PQ to converge to the same point as AB, CD,
cutting the circle in P and P. Draw PO' parallel to PO. Then the
circle through P, with centre O' , is one solution.
in TRIANGLES, CIRCLES AND LINES IN CONTACT 61
Q D,
62 K
62 PRACTICAL PLANE GEOMETRY chap.
64. Problem. To describe a circle which shall pass
through two given points P and Q, and touch a given
circle, centre G.
Join PQ, and draw NK bisecting PQ at right angles.
Select any point C on NK such that the circle, with C as centre
and CQ as radius, will intersect the given circle in E and F say.
Join FE, and produce it to meet PQ produced in R. Through R
draw the tangent RT (or RT'), and produce GT (or T'G) to meet
NK in (or 0') ; then O (or O') is the centre of a circle .satisfying
the given conditions.
For RT 2 =RE.RF=RQ. RP {Euc. III. 36).
Hence the circle through P, Q, T touches RT at T (Euc. III. 37),
and therefore also the given circle at T. Consequently the centre of
the required circle lies on GT produced (converse of Euc. III. 12) ;
but it must lie on NK {Exxc. III. 1) ; hence it must be at O (or G J ).
65. Problem. To describe a circle which shall have
its centre on a given line CD, pass through a given point
P, and touch a given circle, centre G-.
Let fall a perpendicular PN from P on to CD, and produce PN to
Q, making NQ equal PN
Then, by Prob. 64, determine a circle to pass through P and Q,
and touch the given circle, centre G.
66. Problem. To describe a circle to touch two given
circles, centres A, B, and pass through a given point P.
Draw a common tangent Tt to meet AB produced in S ; join PS.
Through G, F, P describe a circle cutting PS in Q.
By Prob. 64 describe a circle to pass through P, Q, and to touch the
circle with centre B ; D being the point of contact. (There are two
such circles.) This circle will also touch the circle with centre A.
Also, by drawing the internal tangent to the given circles, meeting
AB in S', two others may be found by a similar construction.
67. Problem. To describe a circle which shall touch
a given circle, centre Gr, pass through a given point P, and
touch a given straight line AB.
Draw GE perpendicular to AB, meeting the given circle in C and
D ; join CP.
Describe a circle to pass through P, D, E, cutting PC in F.
By Prob. 59 describe a circle passing through P, F, and touching
AB ; this will be the required circle.
Note. There are two circles passing through P, F and touching AB.
Also, by reading D for C, and C for D, in the above instructions,
two other circles will be found to be possible.
in TRIANGLES, CIRCLES AND LINES IN CONTACT 63
64 PRACTICAL PLANE GEOMETRY chap.
68. Problem. To describe a circle which shall touch
three given circles, centres A, B, C.
Let r v r 2 , r 3 denote the lengths of the radii of the circles, centres
A, B, C respectively, the first being the least.
With centre B and radius r., - i\, describe a circle, and with centre
C and radius r 3 - r v describe another circle.
By means of Prob. 66 determine 0, the centre of a circle which
passes through A and touches the two circles just described. Join
OA, OB, OC, meeting the circles at the points indicated. This
construction ensures that AD=EG-FH; from which it follows that
the circle described with O as centre and OD as radius will touch
three given circles externally.
In general there will be eight circles, satisfying the given condi-
tions ; three being such that for each one there is internal contact with
two of the given circles, and external contact with the third ; three
others which have external contact with two of the given circles and
internal contact with the third ; one circle which has internal contact
with the given circles ; finally, the one in the figure.
69. Problem. To describe a circle which shall touch
two given lines AB, AD, and a given circle, centre C.
Draw EF parallel to AD, and GH parallel to AB, the distance
between the parallel lines in each case being equal to the radius of
the given circle.
Now draw a circle which shall pass through C and touch EF and
GH, Prob. 6 1 (there are two such circles) ; let be its centre, Join
OC, cutting the given circle in T.
Then the circle with centre O and radius OT is one solution.
Note. If EF and GH be drawn on the other side of AD and
AB respectively, a second solution will be obtained, the circles having
internal contact.
70. Problem. To describe a circle which shall touch
a given line AB and two given circles, centres C, D.
Draw EF parallel to AB, at a distance equal to the radius of one
of the other circles, say the one with centre C. With centre D describe
a circle whose radius is equal to the difference of the radii of the given
circles.
By Prob. 67 describe a circle to touch the latter circle, pass through
C and touch EF; let O be its centre. Join OC, cutting the circle,
centre C, at G ; then the circle with as centre and OG as radius
will satisfy the required conditions.
Note. There will, in general, be eight solutions. In four cases
EF will lie on one side of AB, and in the other four cases on the
opposite side.
ni TRIANGLES, CIRCLES AND LINES IN CONTACT 6
F
66 PRACTICAL PLANE GEOMETRY chap.
71. Problem. To describe a circle which shall touch a
given circle, centre G, and a given line AB at a given
point T.
1st. Externally. Draw TF z.\\A GD perpendicular to AB.
Join DT, intersecting the given circle at E.
Produce GE to : then O is the centre of the required circle.
2nd. So as to include the given circle. Proceed as above, reading
D', ', and 0' for D, E, and 0.
72. Problem. To describe a circle which shall touch
a given line AB, and a given circle, centre G, at a given
point E.
Join GE, and produce it both ways.
Draw the tangent ER, and set off R T and RT' , each equal to RE.
Draw TO and T'O', each perpendicular to AB ; then and 0'
are the centres of the two required circles, having respectively external
and internal contact with the given circle.
73. Problem. To describe a circle which shall have
its centre on a given line CD, and shall touch a given line
AB and circle, centre G.
By a method illustrating the use of a locus.
Draw any line Jl/JV at right angles to AB, and any line GH,
cutting the given circle in K.
Along MN set off any lengths Ml, Mz, M$, etc., and along KH
set off A'l, A~2, and A'3 respectively equal to these.
With G as centre describe arcs through the points I, 2, 3, etc., on
A'AA to meet lines drawn parallel to AB through the corresponding
points 1, 2, 3, etc., on MN.
These arcs and corresponding lines intersect at points 1, 2, 3, etc.,
through which points draw a fair curve.
This curve is the locus of the centre of a circle which moves so as
always to touch the given line and circle, and the point in which
the curve intersects CD will be such that the circle described with
as centre, to touch AB, will also touch the circle, centre G.
Examples. Take two points A and B 2" apart. Draw a line
CD distant 1" from A and i|" from B. On CD take P, i|"
from A. Describe a circle which shall pass through
1. A and B and touch CD.
2. A, touch CD, and have its centre on AB.
3. A and B, and touch a circle, centre P, radius \" .
4. A, have its centre on AB, and touch circle centre P, radius i".
5. B, touch CD, and a circle, centre A, radius f".
in TRIANGLES, CIRCLES AND LINES IN CONTACT 67
73
68 PRACTICAL PLANE GEOMETRY chap.
74. Miscellaneous Examples.
*1. In the given circle place a chord 1.8" long and parallel to the
line joining the given points /, q. Draw a second chord of the
same length which, if produced, will pass through/. (1884)
Hint. If two circles are concentric, all chords of the outer
circle which touch the inner are equal.
2. Draw two lines including 50 . Describe a circle 2^" diameter
cutting these lines in chords of I-|" and 2|". (1893)
3. Two points a, b are 3|- miles apart. Determine the position
of a point/, such that pa is l miles, and the angle apb 73.
Scale l"=lj miles. (1886)
*4. From a draw a line cutting the two given circles in equal chords.
N.B. An auxiliary curve may be employed. ('893)
Hint. Draw any line through a cutting the circles in chords
ab,cd; along cd mark off ex = ab. Repeat this construction for
other lines through a, and draw a fair curve through the points x.
Let the locus of x cut the circle in X \ then aX is one solution.
"*5. Through A draw two lines containing an angle of 30 , and
intercepting a length of 2" on the given line BC. (1889)
*6. The figure represents a window of the decorated English
style. The construction is sufficiently shown and dimensions
are given. Draw the window to scale of 2" to 1' o". (1879)
7. Describe a circle enclosing and touching three other circles of
0.6", 0.8", and 1" radius respectively, each of which touches
the other two. (1882)
*8. Draw the riband pattern to the given dimensions. (1880)
9. Three posts, B, C, D, are in a straight line at intervals of 100
yards. An observer at A finds that the angle BAC is 20 , and
CAD 30 . Obtain the position of A (Scale TT Vff)- (1882)
10. Construct a regular pentagon having a diagonal of 3". (1881)
11. Draw a triangle from the following data :
(1) Base 2j", perimeter 7 J", one base angle 55 . (1880)
(2) Perimeter 6", vertical angle 6o, altitude 1.25". (1890)
(3) Altitude 2.5", perimeter 8J", one base angle 50. (1877)
*12. A boy starting from a runs in the direction ab. A man starts
at the same moment from c. Supposing the man to run also in
a straight line, but 1^ times as fast as the boy, what direction
must the former take to catch the latter? (1880)
*13. Describe a circle of 2^" radius touching the two given circles.
The given circles are to be within the required circle. (1877)
*14. Describe three circles touching the two given lines AB, CD,
and each other successively
(1) Such that the middle circle passes through P. (1878)
(2) Such that the largest has a radius of 1 inch. (1881)
in TRIANGLES, CIRCLES AND LINES IN CONTACT 69
CHAPTER IV
CONIC SECTIONS
75. Definitions. Let one of two intersecting straight
lines of indefinite length revolve about the other as a fixed
axis, the inclination to each other and the point of intersec-
tion of the lines remaining fixed, then the two equal and
opposite acute angles generate two equal and opposite conical
solid angles of revolution, or a double cone of unlimited
extent. The revolving line generates the surface of the
cone, and in any position is called a generator of the
surface, or simply a generator. The fixed line is called
the axis, and the point of intersection the vertex, of the
cone.
Any plane section of this cone is called a conic section.
The object of the present chapter is to consider the
nature of these conic sections together with some of their
more useful geometrical properties, and to give convenient
methods of constructing the curves.
Classification of conic sections. If the section plane
be at right angles to the axis and do not contain the vertex,
the conic section is called a circle.
If the section plane be not at right angles to the axis,
nor parallel to a tangent plane, and be such that (supposed
indefinitely extended) it cuts only one of the halves of the
double cone, the section is called an ellipse.
If the section plane be parallel to one generator, and
chap, iv CONIC SECTIONS 71
one only, that is, be parallel to a plane which touches the
cone, the conic section is called a parabola.
If the section plane cut both halves of the double cone,
and do not contain the vertex, the section is called an
hyperbola.
If the section plane pass through the vertex, the conic
section will be either a point, a line, or two plane angles
opposite to each other. These may be considered as
limiting cases of the ellipse, parabola, and hyperbola
respectively.
From the manner in which the above conic sections
are formed, it will be obvious that both the circle and
ellipse are completely bounded figures, while the figure of
the parabola extends indefinitely in one direction, and that
of the hyberbola consists of two portions, which extend
indefinitely and in opposite directions. These remarks
refer merely to the general appearance of the sections ; we
have now to consider other relations which distinguish these
figures.
Double meaning of the ivords "cone" and " conic section."
A circle, as defined by Euclid, is the plane figure
enclosed within the circumference, not the circumference
itself, though the word " circle " is frequently used as
synonymous with the latter. The terms "ellipse," "para-
bola," and "hyperbola" are also used in two senses,
denoting in one case a plane figure, and in the other the
bounding curve or outline of the figure. In like manner
the word " cone " is commonly employed with a double
meaning, indicating either the solid figure or its bounding
surface. So the sections of a cone may mean either the
plane figures or their curved outlines.
In this book we shall have occasion to use the terms
in both senses ; the reader must infer the meaning from the
context in any particular case.
In the next article the conic sections are examined from
another standpoint, being regarded as curves traced by a
moving point ; and the nature of the motion is defined.
7 2 PRACTICAL PLANE GEOMETRY chap.
76. Properties of conic sections. Let V be the vertex,
and VO the axis of the cone. DP represents a section
plane which determines the conic section PAQ. Let a
sphere, centre /, be inscribed in the cone so as to touch
the plane at F. This sphere touches the cone in a circle,
centre O, the plane of which {HK) is perpendicular to the
axis ; let this plane intersect the section plane in the line
DN.
Select any point P on the curve PAQ; take PM per-
pendicular to the plane HK, meeting it in M; draw PN
perpendicular to DN, and join PV, PP, ML, MN. In
this way two right-angled triangles PML, PMN are formed,
since PM, being perpendicular to the plane HK, is there-
fore perpendicular to the lines ML, MN in the plane.
Now the angle PLM is equal to the complement of the
semi-vertical angle of the cone, and is therefore constant
for any position of P, hence the ratio PL : PM is constant.
Again, the angle MNP, or a, is equal to angle between
the planes HK, DP and is therefore constant, hence
the ratio PN:PM\s constant,
therefore the ratio PL : PN is constant.
But PL and PP are tangents from P to the same
sphere, therefore PL = PP, hence
the ratio PP: PN is constant.
That is to say, wherever P may be on the curve PAQ, its
distance from F is always in a constant ratio to its distance
from DN, and this is true for the parabola, ellipse, and
hyperbola.
The point F is called a focus, the inscribed sphere a
focal sphere, the line DN is a directrix, and the ratio
PF: /Wis called the eccentricity of the conic section.
Thus we obtain the following definition :
Definition. A conic section is the carve described by a
point which moves in a plane in such a manner that its dis-
tance from a fixed point in the plane (a focus) is in a constant
ratio to its distance from a fixed line (a directrix) in the plane.
IV
CONIC SECTIONS
73
Referring again to the figure, the triangle PLM may-
be supposed to rotate about PM until its plane coincides
with that of the triangle PMN, when LAIN becomes one
straight line. The triangles will then appear as shown
detached at P Q L Q M N. The left-hand diagram of tiie
figure is an elevation on a plane containing the axis of the
cone and perpendicular to the section plane. In these
diagrams corresponding points and angles are denoted by
corresponding letters.
From the manner in which the curves were defined in
Art. 75, it is obvious that for an ellipse the section plane
must be so chosen that a is less than ft, hence P L is less
than jPqNq, or PF is less than PN. For a parabola a is
equal to ft, hence PF is equal to PN. And for a hyperbola
a is greater than ft, and consequently PF greater than PN.
Hence it follows that a conic section is an ellipse, a
parabola, or a hyperbola, according as its eccentricity is
less than, equal to, or greater than unity.
74 PRACTICAL PLANE GEOMETRY chap.
77. The ellipse. Let the cone be cut by the plane
DD' so that the section is an ellipse. A focal sphere is
shown touching the section plane in a focus F, and deter-
mining the directrix FN, as explained in Art. 76. A
second inscribed sphere is shown on the other side of the
cutting plane touching the latter in F'. D ' N' is the line
of intersection of the cutting plane with the plane contain-
ing the circle of contact H'L'K'.
Take any generator meeting the ellipse in P, and the
circles of contact in Z, F. Let" JZ/W be a line parallel
to the axis -of the cone. Draw NPN' perpendicular to
ZW or FN'. Join ML, MN, M 'L ',. MN', FF, FF '.
Suppose the triangles LMP, L'M'F are turned about
MM' into one plane as shown detached to a reduced scale,
then it will be readily seen that
FF : FN= FF' : FN' = PL : FN, or PL' : FN.
Therefore F is similarly related to F' and D ' N' as to F
and FN; thus the ellipse has two foci, F, F', and two
directrices, FN, F'N.
Let the line through FF meet the curve in A, A'.
Bisect FF' in C and draw BB' perpendicular to A A'.
Since the curve may be denned with reference to either
the focus Zand directrix FN, or the focus F 1 and directrix
D ' N', it follows that BB' is an axis of symmetry ; so also
is A A'. Therefore the ellipse has a centre which is C.
Any line through C and terminated by the curve is
called a diameter ; and all diameters are bisected at C.
A A' is called the ma/or axis, and BB' the minor axis ;
they are respectively the longest and shortest diameters,
and are sometimes referred to as the prituipal axes.
It appears from the figure that
PF+ FF' = PL + PL' = LCLC = AF+ AF' = A A'.
Theorem 1. Ln any ellipse the sum of the focal distances
is constant and eqital to the major axis, or to the length of
the portion of a generating line of the cone intercepted by the
circles of contact of the focal spheres.
IV
CONIC SECTIONS
75
L M
Theorem 2. // may be shown that CF, CA, CD are i>i
geometrical progression ; also that the eccentricity FA : AD
= CF: CA = CA: CD.
76 PRACTICAL PLANE GEOMETRY chap.
78. Problem. Having given the lengths of the major
and minor axes of an ellipse, to determine the foci.
Let AA' be the major axis.
Bisect AA' in C ; draw CB, CB' at right angles to AA',
making each equal to half the minor axis.
With B or B' as centre and CA or CA' as radius, de-
scribe arcs cutting A A' in i^and F ' .
Then Fand F' are the required foci.
79. Problem. Having given the foci and the major
axis of an ellipse, to determine the minor axis.
Let F, F' be the two foci and AA' the major axis.
Bisect AA' in C. With Fand F' as centres and radius
CA describe arcs to cut each other in B and B'.
Then BB' is the minor axis.
80. Problem. To construct an ellipse, having given
the major and minor axes AA', BB'. (First method.)
Determine i^and F', the foci, as in Prob. 78.
In CF' take any convenient points 1, 2, 3 . . .
With A 1 as radius describe arcs with centres F and F'
respectively; similarly with A'l as radius describe arcs with
centres F' and F. These arcs intersect in the four points
marked 1,.
Repeat this construction for the points 2, 3, thus obtain-
ing the points marked 2 V 3 r
Then the points i v 2 V 3 X are on the required ellipse.
A fair curve should now be drawn through the points
'i> 2 v 3]/
81. Mechanical method of describing an ellipse.
Insert pins at F, F', B. Take a piece of fine string,
pass it round the pins so as to form a triangle F'BF, and
tie the ends tightly together. Replace the pin B by a
sharply-pointed pencil, with which trace out the curve,
allowing the pencil to be guided by the string, but not
pressing it against the string too tightly.
IV
CONIC SECTIONS
77
B
>
A'l F / 2
3,
3
A **
7
c
A
X
V^j
J
3,
Zi
SI
78, 79, 80
F A
Examples. 1. The lengths of major and minor axes of an ellipse
are 4" and 3" respectively; determine and measure the dis-
tance between the foci. A us. 2.64".
2. The major axis of an ellipse is 3.6" long, and the foci are 2.3"
apart ; find the length of the minor axis. Ans. 2.78".
3. The minor axis of an ellipse is 2.8" long, and the foci are 2"
apart ; find the length of the major axis. Ans. 3.44".
4. Construct the ellipse of Ex. 1 by the method of Prob. 80.
5. Trace the ellipse of Ex. 3 by the method of Art. Si.
j8 PRACTICAL PLANE GEOMETRY chap.
82. Projective properties of the ellipse. It may be
proved that any orthogonal or radial projection of a conic
section is a conic section. It will be sufficient here to state
this theorem in the less general form :
The orthogonal projection of an ellipse is an ellipse.
In this theorem the ellipse is to be considered as in-
cluding the circle and the straight line as extreme cases.
By considering an ellipse as the projection of a circle,
and a circle as the projection of an ellipse, we shall be
able to prove in a simple manner some useful properties of
the curve. In this connection two additional theorems of
projection will be required.
A system of parallel straight lines project orthogonally
into a system of parallel straight lines, the lengths of the pro-
jections bearing the same ratios to each other as the lengths of
the corresponding lines bear to each other.
If a line be tangential to a curve at any point, then any
projection of the line will be tangential to the projection of the
curve at the projectio?i of the point.
Let AA', B Q B ' be two diameters of a circle perpen-
dicular to each other.
Suppose the circle to be turned about AA' as axis until
the projection of B B Q ' on the plane of the paper is BB' ;
then the projection of the circle will be the ellipse shown,
of which AA ', BB' are the principal axes.
Next conceive the ellipse, the plane of which is that of
the paper, to be turned about the minor axis until the major
axis projects into the line A X A^ of length equal to BB' ;
then the projection of the ellipse will be the circle with
BB' as diameter.
Definitions. The circle described on the major axis
of an ellipse as diameter is called the major auxiliary
circle, or simply the auxiliary circle or the major circle.
The circle with the minor axis as diameter is called the
minor auxiliary circle or the minor circle.
If B be any point on the major auxiliary circle, then
when this circle is turned about AA' and projected into
IV
CONIC SECTIONS
79
T '' i
%C ' s
B
o/\\
< GL^^Ti
S /
r\ M\
\
^fT^V
1 (a: \
A\\
\
R
b lA
^
C
B'
\M
\A
the ellipse, the projection of P Q will be P, where P P is
perpendicular to CA. And from the principles of projection
CA : CB = constant
P M: PM=P C: BC--
for all points on the curve.
Again, if the ellipse be turned about the minor axis
and projected into the minor auxiliary circle, the point P
projects into P x , where PP 1 is perpendicular to CB and
PN\ P X N= AC: A 1 C=CA: CB
P M: PM.
From this relation it is easy to prove that the line P P 1
if produced will pass through C.
These theorems may be stated as follows :
Theorem i. The ratio of the ordinate of any point of
an ellipse to the corresponding ordinate of the major auxiliary
circle, is constant and equal to the ratio of the mi?ior to the
major axis. And this is equal to the inverse ratio of the
abscissa of the point to the abscissa of the correspo?iding point
on the minor auxiliary circle.
See Art. 144 for definitions of ordinate and abscissa.
So PRACTICAL PLANE GEOMETRY chap.
Theorem 2. If P be any point on an ellipse, ike
corresponding points P and P x on the auxiliary circles tie on
the same radius.
Next, let the tangent at any point Q on the ellipse
intersect the major and minor axes in R and S. If
the ellipse, with the tangent, be turned about the minor
axis so as to project into the minor auxiliary circle, then Q
will come to Q v S will not move, and the tangent SQ to
the ellipse will project into the tangent SQ 1 to the minor
auxiliary circle. Similarly RQ will be the tangent to the
major auxiliary circle corresponding to tangent RQ to the
ellipse. And RQ , SQ l are parallel to each other since
<2 , <2i ue on tne same radius ; we thus have the following
theorem :
Theorem 3. A tangent to an ellipse and the corre-
sponding tangent to the auxiliary circle intersects the major
axis at the same point. Also the tangent to the ellipse and
the corresponding tangent to the minor auxiliary circle
intersects the minor axis at the same point. And the two
auxiliary tangents are parallel to each other.
Bisect P X P in O and join OP. Then it is clear that
for all positions of the point P
nn CP, + CP n CA + CB
CO = - 1 = = constant.
2 a
and also that
CA - CB
OP = OP 1 = OP = = constant.
Further, since both the triangles PfDP, P Q OP are
isosceles, it is evident that CO and OP are always equally
and symmetrically inclined to CA ; they are also equally
and symmetrically inclined to CB. We are thus led to the
following theorem :
Theorem 4. If two lines CO and OP of constant length
move round a fixed point C so as to be always symmetrically
inclined to a given fixed line, the locus of P will be an ellipse
IV
CONIC SECTIONS
of which the principal axes are lines through C parallel and
perpendicular to the fixed line, the lengths of the semi-axes
being respectively equal to the sum and difference of CO and
OP.
A mechanism for draining ellipses has been constructed
on this principle. CO is a crank free to rotate on a pin
fixed at C to a frame. OP is a second crank pinned
to the first one at O, and constrained by mechanism to
turn (i.e. alter its angular position) at the same rate that
CO turns, but in the opposite direction. In order to be
able to trace an ellipse of any given size, it is necessary that
the lengths of both cranks be capable of adjustment, one
being made equal to half the sum, and the other to half
the difference of the semi-axes. The figure shows OP as,
the shorter crank, but the same ellipse would be traced if
the lengths of CO and OP were interchanged so that OP
were the longer.
Example. Construct the above figure when the lengths of the
axes AA', BB' are 4" and 2j".
G
82 PRACTICAL PLANE GEOMETRY chap.
83. Problem. To construct an ellipse, having given
the major and minor axes. (Second method.)
Let A A' and BB' be the major and minor axes.
With centre C describe the two auxiliary circles on A A ' ,
BB' as. diameters.
Divide the arc AB into a number of equal parts,
say six ; draw the radii to the points of division, meeting
the inner circle in corresponding points. Draw lines
through the outer set of points perpendicular to the major
axes to meet parallels thereto through the inner set.
A fair curve must be drawn through the points thus
found. The remaining portions of the ellipse may be
similarly determined, or obtained by constructions depend-
ing on symmetry.
84. Problem. Having given the principal axes of an
ellipse, to find points in the curve mechanically by means
of a paper trammel.
Let AA', BB' be the two axes. Take a strip of paper
with one edge straight, and from a point P on this edge
mark off Pa, Pb, equal to CA, CB, the semi-axes.
Place the strip in successive positions with a and b
always on the minor and major axes respectively, and mark
corresponding positions oi P \ these will be points on the
ellipse.
For, bisect ab in O and join CO. Then
CO = Ob = \ab = constant
and angle OCb = angle ObC.
Hence CO, OP, two lines of constant length, move round
C, and in all positions are symmetrically inclined to a fixed
line CA ; therefore by Theorem 4, Art. 82, the locus of P
is an ellipse of which
the semi-major axis = PO + CO = PO + Oa
= Pa = CA
and the semi-minor axis = PO - CO = PO - Ob
= Pb = CB.
An instrument called a trammel for drawing ellipses is
constructed on this principle.
IV
CONIC SECTIONS
3
If Pa and Pb had been set off opposite ways along the
edge of the trammel, the same ellipse would have been
described. CO would then have been the longer of the
two cranks.
Xotc. Tracing-paper may with advantage be used for the trammel,
in which case the points /*, a, b need not be on its edge.
84 PRACTICAL PLANE GEOMETRY chap.
85. Problem. A triangle moves in such a manner
that two of its angular points always lie respectively in
two fixed lines, to determine the locus of the third
angular point.
Let PQR be one position of the triangle, the points Q
and R being on the fixed lines CVand CX respectively ;
the locus of P is required.
Describe a circle to pass through C, Q, R, and let O be
its centre; join and produce PO, cutting the circle in b and a.
Draw the lines CaB, CbA, and make CA = Pa and CB
Pb. The locus of P will be an ellipse of which CA and
CB are the major and minor semi-axes.
For, join OR, and suppose the circle and the lines
PO, OR to be drawn on the plane of the triangle and to
move with it.
The radius of the circle, the chord QR, and the angle
QCR remain constant in magnitude, therefore as the triangle
and circle move together the latter always passes through
C (Euc. III. 21). Hence
CO moves about C as centre . . . (i.)
Now suppose that the triangle turns through any angle 0,
say clockwise; then OR will do the same (since it is
attached to the triangle), and the angle ORC will be
increased by an amount 6. But since OCR is isosceles
the angle OCR will also be increased by an amount 6, and
therefore OC will turn through an angle anti-clockwise.
Hence
CO and OP turn about C and O respectively
at the same rate but in opposite directions . (ii.)
But in the position shown in the figure, CO and OP are
symmetrically inclined to the fixed line CA, since OCb is
isosceles, therefore from (ii.)
CO and OP turn about C and are always
symmetrically inclined to the fixed line CA (iii.)
Hence from (iii.), as stated in Theorem 4, Art. 82, it
follows that the locus of P is an ellipse of which
IV
CONIC SECTIONS
35
c
the semi-major axis = PO + OC '= Pa = CA
and the semi-minor axis = PO - OC Pb = CB,
the directions of which are CA and CB.
It will be noticed that Pba is the trammel for the double
crank CO, OP.
The triangular trammel PQP, the double crank CO, OP,
and the principal straight line trammel Pba are thus seen to
be equivalent to one another in determining the path of P.
An indefinite number of equivalent trammels could be
found, either triangular or straight ; for the former by taking
any two fixed lines CX\ CY', intersecting the circle in
R', Q', and joining PQ', PR' ; or for the latter by drawing
any line through P intersecting the circle in a, b', and then
taking Ca', Cb' as the fixed lines.
Examples. 1. The lengths of the major and minor axes of an
ellipse are respectively 4" and 3". Set out the curve (a) by the
method of Prob. 83 ; (6) by the method of Prob. 84, using
tracing-paper for the trammel.
2. Work Prob. 85, having given
Angle XCY=-jo, QP=2.$", QP=2.-j$", RP= 1.0".
Measure the lengths of the principal axes of the ellipse described
by P, and the angle which the major axis makes with CA'.
Ans. 6.55", 1.27", 13. 7 .
86 PRACTICAL PLANE GEOMETRY chap.
86. Conjugate diameters of an ellipse.
Let tangents be drawn to a circle at the ends of two
diameters perpendicular to each other, thus forming a
circumscribing square as shown in the figure ; and let PQ,
PR be any two chords of the circle respectively parallel
to the diameters.
Suppose this figure to be turned about any line XX
so that its plane is inclined to the plane of the paper,
and the northogonally projected ; the shape of the projec-
tion will be as shown to the right, corresponding points
being denoted by corresponding letters.
From the principles of orthogonal projection we infer
that the square with the inscribed circle touching it at the
middle points of its sides will project into a parallelogram
with an inscribed ellipse touching the sides at their middle
points. It also follows that </and kg, the tangents at the
ends of the diameter jf, are parallel to the diameter i'i ;
similarly the tangents eh and fg at the ends of the diameter
i'i are parallel to the diameter jf.
And since the chords PQ and PR are bisected by JJ'
and 77' at M and irrespectively, so also are the chords/^
and pr bisected by jj' and ii' at m and n respectively.
Such diameters of an ellipse as ii' and^/' are said to be
conjugate ; the tangents to the ellipse at the ends of either
diameter are parallel to the other diameter ; each diameter
bisects all chords parallel to the other.
87. Problem. Having given a pair of conjugate
diameters of an ellipse, to determine the principal axes.
Let 77', JJ' be the given conjugate diameters. Draw
JG perpendicular to and equal to CI. Join CG, and on
CG as diameter describe a circle, centre O. Join JO,
intersecting the circle in b and produced in a. Join Cb,
Ca, and on these lines produced take CA and CA', each
equal to Ja, and CB, CB ', each equal to Jb.
Then AA' and BB' are the major and minor axes of
the ellipse.
IV
CONIC SECTIONS
87
For, join if, then by comparing with Prob. 85 it is seen
that Jij is the triangular trammel which will give the re-
quired ellipse ; CO, OJ is the double crank, and Jba the
principal straight line trammel equivalent to the triangular
trammel in Prob. 85.
Example. The lengths of two semi-conjugate diameters of an
ellipse are 4" and 3", and the included angle 6o 3 ; determine
the principal semi-axes, and the angle which the major axis
makes with the larger conjugate diameter.
A ns. 4.40", 2.35", 1 7. 5 ,
88 PRACTICAL PLANE GEOMETRY chap.
88. Problem. To describe an ellipse, having given a
pair of conjugate diameters, or to inscribe the principal
ellipse in a given parallelogram.
Let DEFG be the given parallelogram, or II', Jf the
given conjugate diameters.
First Method. On two adjacent sides of the parallelo-
gram, DG, DE, as diameters, describe semicircles, and
divide each into the same number of equal parts, say six.
From the points of division draw lines perpendicular to
the sides, intersecting the latter in points marked i, 2.
From the points on DG draw lines parallel to DE ; and
from the points on DE draw lines parallel to DG.
These two series of lines will intersect in a series of
points i p 2j on the required ellipse.
Second Method. Divide CI and EI each into the same
number of equal parts, say three, and draw lines from /'
and /to the points 1, 2, as shown.
The intersections of corresponding lines will give points
1 2 j, which are on the required ellipse.
Proof. The above two constructions are readily seen to be true for
a square with inscribed circle, of which the figure may be regarded
as the projection. Hence they are true for the projection.
Third Method. By a triangular trammel.
Draw JjG x perpendicular to CI, and make JG 1 = CI.
Draw / perpendicular to Jf and join ij. Then Jji is the
shape of the required triangular trammel.
Take a piece of tracing-paper for the trammel and mark
the three points, /, /, / on it ; then if i and / move on JJ'
and W respectively, P (or /) will trace the ellipse.
It will be noticed that the angle ijj (or iPj) is equal to
the complement of the angle ICJ, and that the lengths of
Pi and Pj are respectively equal to the perpendicular
distances of /and /from the conjugate diameters.
Proof. The point G x is the instantaneous centre of rotation of
the trammel when moving through the position shown ; hence the
ellipse described by P touches DE at/. It can be shown in a similar
manner that the ellipse touches EF sX I.
IV
CONIC SECTK )NS
89
/'..-
88
89. Problem. Having given the curve of an ellipse,
to determine the principal axes. (No figure.)
First Method. Draw any two chords, PP', QQ', parallel
to each other. Draw the line bisecting these chords and
meeting the curve in /, I', then IP is a diameter, and C,
its middle point, is the centre of the ellipse.
Through C draw the diameter JJ' parallel to PP' .
Two conjugate diameters are thus determined, and the
major and minor axes can be found as in Prob. 87.
Second Method. Proceed as in the first method so far
as to find C.
Then with centre C and any suitable radius, describe a
circle cutting the ellipse in P, Q, P, S (taken in order).
Draw the diameters PP, QS, and bisect the angles
between them. The bisecting lines terminated by the
curve are the principal axes.
Examples. 1. Two conjugate diameters of an ellipse are 4-^" and
3V' long and make 6o with one another ; describe the ellipse
by each of the three methods of Prob. 88.
2. Rule any two intersecting lines in ink on drawing-paper. Mark
any three points on thick tracing-paper. Let two of the points
move one on each line, and prick off points on the ellipse traced
by the third point. Draw a fair curve through the points.
Find the principal axes of the ellipse by each of the methods of
1'rob. 89. Confirm the result by the construction of Prob. 87.
9 o PRACTICAL PLANE GEOMETRY chap.
90. Tangent and normal to any plane curve. Curvature.
Let P, Q be two points in any given curve, and
suppose the point Q to move on the curve and to approach
indefinitely near to P, then the chord PQ will turn round
P and tend to a definite direction PT, becoming in the
limit the tangent to the curve at P. The line PG, per-
pendicular to PT, is called the normal to the curve at P.
Definitioti i. The tangent at a point in a curve is the
straight tine passing through the point and through a second
point in the curve indefinitely near the first. The normal to
the curve at the point is the straight line passing through the
point at right angles to the tangent.
This general definition of a tangent does not indicate a practical
method of accurately drawing it at any given point in the curve. In
order to be able to draw the tangent in the right direction it would be
necessary to know something more with regard to it, such as the
position of a second point in it at a finite and sensible distance from
the first point (or a line to which it is parallel, perpendicular, or
inclined at a known angle, or a second curve which it touches, etc.),
and this would require special knowledge of the particular curve under
consideration.
If it be required to draw a tangent from a point R not in the given
curve, this may be done with all attainable precision, by applying a
straight-edge, adjusting its position and then drawing the line RS to
touch the curve as shown. The exact point of contact at S would be
still undetermined, and if this point, as well as the line of the tangent,
were required, an additional construction would be necessary, such as
the drawing of a line intersecting the tangent at the required point.
This additional construction would require to be based on some known
property of the given curve.
Let Q, P, R be any three points on a plane curve ;
bisect the two chords QP, PR at right angles by the lines
AC, BC meeting at C ; then a circle with centre C passing
through P will also pass through Q and R. Suppose now
the two points Q and R to move along the curve and to
approach indefinitely near to P. Then in the limit the two
bisectors AC, BC become normals to the curve; they
intersect in a definite point O, called the centre of curvature.
The circle through P with centre O is called the circle of
curvature at P, and OP is the radius of curvature.
IV
CONIC SECTIONS
91
Definition 2. The circle of curvature at any point in a
curve is the circle passing through the point and through two
other points in the curve indefinitely near the first. Its radius
is the radius of curvature, and its centre, called the centre of
curvature, is the point of intersection of the normal at the
point on the curve, with a second normal at a point indefinitely
near the first.
The centre and radius of curvature cannot be accurately determined
by directly applying the above construction ; a special construction is
required, depending in each case on the nature of the curve.
Any circle through P with its centre on the normal at P would
touch the curve, the two would have a common tangent, or have two
consecutive points in common ; but the circle of curvature coincides
more closely with the curve at the point than any other circle which
can be drawn ; as we have seen, it has three consecutive points in
common with the curve. On examining the figure it will be noticed
that the circle of curvature crosses the curve at P ; on one side of P
the curve gradually becomes flatter, and so falls away from the circle
on the outside ; on the other side of Pihe. curvature gradually increases,
and so the curve bends away from the circle towards the inside. It
is thus seen that the amount of curvature at P is correctly measured
by the circle of curvature ; and the latter always crosses the curve
unless the curvature is a maximum or minimum, as, for example, at
the ends; of the principal axes of an ellipse ; in this case the curve and
the circle of curvature havefo/tr consecutive points in common.
92 PRACTICAL PLANE GEOMETRY chap.
91. Focal and tangential properties of the ellipse.
Let PT, PG be the tangent and normal to the ellipse
at any point P on the curve.
Draw FH, F H' perpendicular to the tangent. Draw
the focal lines FP, FP, and produce one of them.
Then it may be shown that the angle FPG = angle
FPG, and that the angle FPT= angle APT. Also that
ar=cn'=CA. Or
Theorem i. The normal at any point P in an ellipse
bisects the ang/e between the focal lines PF, PF, and the
tangent bisects the angle between one of the focal lines and
the other one produced.
Theorem 2. The feet of the perpendiculars from the
foci to any tangent to an ellipse lie on the major auxiliary
circle.
92. Problem. To draw the tangent and normal to a
given ellipse at a given point on the curve.
First Method. Let P be the given point. Join FP,
F'P; produce one of these, say F'P to A', and bisect the
angle FPA'by the line PT; then PT is the tangent at P.
Draw PG perpendicular to PT, or bisect the angle
FPF; then PG is the normal at P.
Second Afethod (refer to figure on p. 95).- Let Q be
the given point. Determine Q , the corresponding point
on the major auxiliary circle. Draw the tangent Q P,
intersecting the major axis in P ; join RQ ; then PQ is the
required tangent.
Or if P be too remote, determine Q v draw the tangent
Q-yS; then SQ is the required tangent.
The normal at Q is not shown in the figure, but it
passes through Q, and is perpendicular to PS.
Third Method, (see Fig. 84). Let P be the given point.
With centre P and radius equal to CA, describe an arc
intersecting the minor axis in a. Join Pa, intersecting the
major axis in b. Through a and b draw lines (not shown)
perpendicular to axes BB ', A A', intersecting in G ; then
PG is the required normal.
IV
CONIC SECTIONS
93
For G is the instantaneous centre of the trammel Pba, and therefore
the tracing-point P is moving in the direction at right angles to PG.
93. Problem. Having drawn a tangent to a given
ellipse, to find the point of contact.
First Method. Let the tangent be the one shown in the
figure above. From one focus T^draw FHK perpendicular
to the tangent intersecting it in H, and make HK=FH.
Join KF' intersecting the tangent in F ; then P is the
point of contact.
Second MetJiod (see the figure on p. 95). -Let the tangent
intersect the major and minor axes in R and S.
From R draw a tangent RQ to the major auxiliary circle,
and determine the point of contact Q by drawing CQ
perpendicular to RQ - Draw QQ perpendicular to A A'
to meet the tangent in Q. Then Q is the required point
of contact.
If the point R is not available, make the corresponding
construction for the minor auxiliary circle, i.e. draw SQ V
CQ V Q X Q-
Example. Draw the above figure one-half larger.
94
PRACTICAL PLANE GEOMETRY chap.
94. Problem. Having given the principal axes (or the
foci and one axis) of an ellipse, to determine the two
tangents to the curve from a given external point (with-
out drawing the curve), and to find the points of contact.
First Method (figure, last page). Let T be the given
point. Join T to one of the foci, say F. On TF as
diameter describe a circle, and draw also the auxiliary
circle on AA'. Let these circles intersect in B, H x ;
then TJ7, TH X are the required tangents.
To find the points of contact, produce FH to K, and
make HK=HF. Join KF, intersecting the tangent in
F; then F is the required point of contact for one of the
tangents ; that for the other is found in a similar manner.
Second Method (figure opposite). This method is based
on the projective properties relating to the ellipse and the
auxiliary circles. See Art. 82.
First let the given point be R, on the major axis.
Draw RQ to touch the major circle, and draw CQ
perpendicular to FQ , thus determining the exact point of
contact <2 , and also the point Q x on the minor circle.
Through Q Q , Q x draw the lines parallel to the axes, meet-
ing in Q. Then RQ is a tangent, and Q the point of contact.
Next let the given point be S, on the minor axis.
Draw SQ V tangent to the minor circle, then draw the
perpendicular CQ X Q^ thus determining Q v Q , from which
Q may be found as in the last case.
Lastly, let the given point be T, on neither axis.
Draw TL perpendicular to AA'. Make LD = LT.
Draw DT parallel to A X 'B (or perpendicular to AB\
intersecting LT produced in T v Then T is a point in
the auxiliary tangent to the major circle.
For 7' Z : TZ = B C : FC= Q F : QE.
This tangent T Q is then drawn, and the points Q , Q v Q
found as before. The required tangent from T is TQ, and
Q is the point of contact.
The second tangent would correspond with the second
tangent (not shown) from T Q to the major circle.
IV
CONIC SECTIONS
95
Examples. 1. Draw the above figure double size.
2. Using tracing-paper for a trammel, construct an ellipse whose
principal axes are 4^" and 3" long. Select any point P on the
curve, and determine the tangent and normal at P by each of
the methods of Prob. 92.
3. In Ex. 2 select any point Q external to the ellipse, and through
Q draw a tangent to the curve. Then determine the point of
contact by each of the methods of Prob. 93.
4. Draw two lines AA ', BB ', 4" and 3" long, bisecting each other
at right angles in C. Along CA, CB produced, set off CR,
CS equal to 3.6" and 2.8"; and mark a point T such that
AT=.f, BT=2.7".
By each of the methods of Prob. 94 draw from R, S, and T
tangents to the ellipse of which AA', BB' are the principal axes,
without drawing the curve, and find the points of contact.
5. Find the eccentricity of the ellipse of Ex. 2, and the distance
apart of the directrices. Ans. 0.745: 6.04".
Hint. See Theorem 2, Art. 77.
6. Having given one focus F and two tangents TP, TP' with their
points of contact P, P' ; to find the other focus F' .
Hint. See Fig. 91. Draw FHK and join KP. Similarly
obtain KP' . Then F' is at the point where KP and K ' P'
meet.
96 PRACTICAL PLANE GEOMETRY chap.
95. Problem. Having given the major and minor
axes of an ellipse, to find, without drawing the ellipse,
the points (if any) in which a given line intersects the
curve.
The method adopted is based on the projective pro-
perties of the auxiliary circles described in Art. 82.
Draw the auxiliary circles on the given axes AA' and
BB' as diameters. Let the given line intersect the major
auxiliary circle in L, M.
Join CL, CM, cutting the minor auxiliary circle in /, m.
Through L, M draw lines parallel to A A' to meet in L v M 1
lines through /, /// drawn parallel to BB'.
Join L X M X intersecting the minor auxiliary circle in P v
Q v Through P v Q x draw lines parallel to AA' to meet
LM in P, Q\ then P and Q are the required points of
intersection of LM and the ellipse.
If the given line be perpendicular to the major axis, let
H be the point in which it meets the major circle ; join
CH, cutting the minor circle in D , and draw DAI perpen-
dicular to BB' to meet the given line in D ; then D is one
of the required points of intersection.
If the given line be perpendicular to the minor axis, let
K be the point in which it meets the minor circle. Join
CK and produce it to meet the major circle in P ; draw
E Q E perpendicular to AA' to meet the given line in E ;
then E is one of the required points of intersection.
96. Problem. To describe an ellipse having given its
centre C and three points P, Q, R on the curve.
Join CQ and. PR intersecting in H. With centre C
radius CQ, describe a circle, and through Zf draw the chord
pr of this circle, such that/ZT: Hr = PH\ HR ; and draw
the radius Cs perpendicular to CQ. Join sr (or sp) inter-
secting CQ in A". Join and produce RK (or PR) to meet
in S a line drawn through s parallel to Rr or Pp, and join
CS.
Then CS, CQ, are conjugate semi - diameters of the
IV
CONIC SECTIONS
97
96
SyS.
required ellipse. The principal axes can be determined
by Prob. 87.
This construction is based on some theorems of projection. The
required ellipse is supposed to be projected into a circle. A figure
similar to the projection is drawn of such a size and so placed that CQ
coincides with its own projection. The first part of the construction
locates / and r, the projections of P and R. The second part locates
S, a point which projects into s, where Cs has been drawn conjugate
to CQ in the projection.
H
98 PRACTICAL PLANE GEOMETRY chap.
Examples. 1. In Ex. 4, p. 95, join RT. Also draw a line DD
parallel to A A' at a distance of .5" ; and a line EE parallel to BB'
and distant .7" therefrom. By the method of Prob. 95 determine the
points in which the lines RT, DD, EE would cut the ellipse, if the
latter were drawn.
2. Work Prob. 96, taking for the data CP=l.$" i CQ=l.f, CR
= 2.2"; PCQ=72, QCR=34 a . Complete the figure by drawing
the ellipse through P, Q, R.
97. Problem. To construct an ellipse, having given
the foci F, F' and a tangent.
(See figure, Art. 91.) From one of the given foci, F,
draw a perpendicular to the given tangent meeting it in H;
then H is on the major auxiliary circle.
Bisect FF' in C, and with centre C and radius CH
describe the circle cutting FF' produced in A, A' ; then
AA' is the major axis.
The minor axis may now be found as in Prob. 79, and
the ellipse constructed by one of the methods already given.
98. Problem. Having given one axis of an ellipse
and a tangent, to construct the curve.
If the major axis be given, describe the auxiliary circle
on it, and from the two points in which the given tangent
cuts the circle, draw lines perpendicular to the tangent, to
meet the major axis in F, F', the foci (see figure, Art. 91).
The minor axis can now be found and the ellipse con-
structed.
If the minor axis be given, describe the minor circle on
it and let .S (figure, page 95) be the point in which the
given tangent cuts the given axis (produced if necessary).
Draw the tangent from S to the circle, and from C draw
CQ X perpendicular to the tangent.
Draw Q x Q perpendicular to BB' to meet the given
tangent in Q, and from Q draw QQ parallel to BB' to
meet CQ X produced in Q Q ; then Q is a point on the
major auxiliary circle, and hence the major axis is known
and the ellipse can be constructed.
IV
CONIC SECTIONS
99
THE PARABOLA
99. Problem. To construct a parabola, having given
the focus F and directrix DD'.
It has been explained in Art. 75 that the section of a cone by a
plane parallel to a tangent plane is a parabola. And in Art. 76 it was
shown that this curve may also be denned as the locus of a point
which moves in a plane in such a manner that its distance from a
fixed point, the focus, is always equal to its distance from a fixed line,
the directrix ; this property enables us to set out the curve.
Through i^draw a line perpendicular to DD\ meeting
the latter in Z. Bisect ZF in A.
Then A is on the parabola, for AF= AZ. A is called
the vertex, and the perpendicular ZF the axis of the para-
bola. The latter is a line of symmetry for the curve.
To draw the curve. Select a series of points 1, 2, 3 . . .
on the axis as shown, and through these points draw lines
perpendicular to the axis.
With F as centre, Z\ as radius, describe an arc to inter
sect the perpendicular through 1 in i r
Repeat this construction for the other points, and draw
a fair curve through the points A, i p 2 p 3 X . . . thus
found.
ioo PRACTICAL PLANE GEOMETRY chap.
100. Properties of the parabola. In the upper figure
PN and PM are perpendicular to the axis and directrix.
PT is the tangent and PG the normal at P, and A Y is the
tangent at the vertex. PN is called the ordinate, and NG
the subnormal for the point P. The double ordinate LL X
through the focus is called the latus rectum.
In the lower figure let PQ be any chord of a parabola;
PT, QT the tangents at P and Q ; TN a line parallel to
the axis, intersecting the curve at A ; and let XY be the
tangent at A.
The following theorems are proved in works on pure
mathematics. The student should commit them to
memory, and test them by setting out the figures to scale.
Theorem i. The tangent PT bisects the angle PPM.
Theorem 2. A line drawn through the focus perpendicular
to a tangent, meets the latter at a point which is on the tangent
at the vertex.
Theorem 3. AT and AN are equal. (In both figures.)
Theorem 4. The latus rectum LL X is equal to 4AP.
Theorem 5. The length of the subnormal NG is constant,
and equal to half the latus rectum, or to 2AP, or to PZ.
Theorem 6. PN 2 is proportional to AN. Or the square
of the ordinate is proportional to the abscissa. For the upper
figure PN 2 = Z l -AN=4AF-AN Also for the lower
figure PN 2 ^AF- AN, where Pis the focus.
Theorem 7. A line which bisects any system of parallel
chords is called a diameter, and is parallel to the axis, which
is a particular diameter.
Theorem 8. The line TN, through T, parallel to the axis,
bisects the chord PQ, and TN is bisected at A. That is, PN
= NQ, and TA = AN
Theorem 9. The tangent XY at A is parallel to PQ, and
XY= \PQ.
Theorem 10. Tangents which meet in the directrix are at
tight angles to each other.
Theorem 1 1. The area ALPNA is two-thirds the circum-
scribing rectangle AN- PN.
IV
CONIC SECTIONS
IOI
101. Problem. To draw the tangent and normal
to a given parabola at a given point P on the curve.
(See the upper figure.) Join P to the focus F, and draw
PM parallel to the axis ; bisect the angle PPM by the
line PP. Then PP is the tangent at P.
Or, make PP= PP and join PP.
Or, make AP=AJV and join PP.
To draw the normal, first draw the tangent as above,
and then draw the normal PG perpendicular to it.
Or, make NG = PZ, and join PG.
io2 PRACTICAL PLANE GEOMETRY chap.
102. Problem. To draw the tangent to a parabola
from an external point T.
Join T to the focus F, and on TF as diameter describe
a circle to cut the tangent at the vertex in H and H' .
Then FH, TH' are tangents to the parabola.
103. Problem. To describe a parabola, having given
two tangents with their points of contact.
Let FT, QT be the given tangents, intersecting in T,
and let F, Q be the given points of contact.
Join FQ and bisect FQ in N Join FN and bisect
FN in A. Draw through A a line Rt parallel to QF.
Then 7Wis parallel to the axis of the parabola, A is a
point on the curve, and FA is a tangent at A.
First Method. Shown on the left of FN. Draw QR
parallel to NA, so that QRANis a parallelogram.
Divide NQ and RQ into any and the same number of
equal parts, say four. From the points of division i, 2, 3
on NQ draw lines parallel to NA ; and from the points of
division 1, 2, 3 on RQ draw lines through A.
Let the parallel lines intersect the corresponding radial
lines in the points I. II. III., as shown. A fair curve
through these latter points is the parabola required.
The curve may be extended beyond Q by continuing
the divisions 5, 6, ... on NQ, RQ, and through them
drawing the parallels and radials, intersecting in V. VI. . . .
The same construction might be used for the portion of
the curve AF. Instead we give the following :
Second Method. Shown on the right of FN Deter-
mining A as before. Let the tangent at A, which is
parallel to QF, intersect FF'm t.
Join FA, and through / draw a line parallel to AN,
intersecting FA in n. Bisect hi in a. Then a is a point
on the curve, and the tangent at a is parallel to FA.
This construction for determining a is the same as the
one that was used for finding A. If desired, further points
between Aa and aP can be found by repeating the process.
IV
CONIC SECTIONS
103
104. Examples on the parabola.
1. Construct a parabola for which the distance of the focus from
the directrix is ". Measure the latus rectum. Ans. ij".
2. In Ex. 1 take a point P on the curve distant 2" from the focus,
and draw the ordinate, the tangent, and the normal at /'.
Measure the length of the subnormal. Ans. ^".
3. In Ex. 1 select any point external to the parabola, and draw a
tangent to the curve. Determine the point of contact.
4. Draw a triangle TPQ, having given PQ-3", PT= 2f", QT= 2".
By each of the methods of Prob. 103 construct the parabola
which touches PT and QT at /'and Q.
5. Determine the axis, vertex, focus, and latus rectum of the para-
bola of Ex. 4. Ans. Length of the latus rectum = 2.24".
6. Draw any parallelogram, and in it inscribe a parabola which
touches one side at its middle point, and passes through the
ends of the opposite side. Determine the latus rectum.
7. A jet of water issuing from a horizontal nozzle strikes a point I
foot below the orifice and 3 feet distant horizontally. Draw
the path of the jet ; scale x \. Find the latus rectum of the
path, and obtain the horizontal velocity of the water. Ans. 9
feet ; 12 feet per second.
Hint. It is known that (neglecting air resistances) the path
of an object moving freely under the action of gravity is a para-
bola, with axis vertical and latus rectum in feet equal to z> 2 -r 16,
where v is the horizontal velocity in feet per second.
104 PRACTICAL PLANE GEOMETRY chap.
THE HYPERBOLA
105. Properties of the curve. "\Ye explained in Art.
75 that a hyperbola was obtained when a cone was cut by
a plane so taken as to penetrate both portions of the com-
plete surface ; and in Art. 76 it was shown that the curve
might be defined as the locus of a point which moves, so
that its distance from a fixed point bears a constant ratio
(greater than unity) to its distance from a fixed line.
The curve is set out to scale in the figure ; it is seen to
consist of two parts detached from each other, each part
having two branches ; these branches extend to infinity on
the complete curve.
Although so different in appearance from the ellipse,
the two curves have many closely allied relations. Both
have two axes of symmetry, two foci, two directrices, and
each has a centre. Each curve, however, has properties
peculiarly its own.
The hyperbola is not treated as fully in this work as is
the ellipse, on account of its minor importance in the arts.
We shall define the terms, state some of the properties,
and give the more useful problems. The student should
illustrate the subject by drawings to scale. Proofs may be
found in works on pure mathematics.
Definitions. The points F, F' are the foci.
C is the centre, chords through which are bisected at C.
A and A' are the vertices, and AA' is the transverse axis.
BB ', perpendicular to AA', is called the conjugate axis,
being determined by the condition that AB = AB' = CE
= CE' = CF.
The diagonals DE', D ' E, produced indefinitely both
ways, are asymptotes to the curve.
Asymptotes are lines which, as they recede to infinity with
a curve, approach nearer a?id nearer to the latter without limit,
but never actually coincide with it.
Properties. Let P be any point on the curve. [oin P
to Fznd F' ; then P'F- PF= A A' ; or
IV
CONIC SECTIONS
105
Theorem 1. The difference of the focal radii is constant,
and equal to the transverse axis.
Observe that in the ellipse it is the sum which is constant.
Theorem 2. The tangent PT at any point P bisects the
focal radii PP, PP' .
In the ellipse it is the normal which bisects this angle.
From any point Q on the curve, draw lines parallel to
the asymptotes, meeting the latter in M or N. Then
Theorem 3. The product QM QN is constant.
This important property is characteristic of the hyperbola.
Let X be the point where a directrix cuts CA (not
shown), then it may be shown that
Theorem 4. The eccentricity FA : AX CA : CX
= CP: CA, and CX, CA, CF are in geometrical progression.
Compare Theorem 2, Art. 77, for the ellipse.
io6 PRACTICAL PLANE GEOMETRV chap.
106. Problem. To construct a hyperbola, having given
the focii F, F', and the transverse axis AA' (Fig. 105).
Select points 1, 2, 3, 4 . . . on the axis, outside the
foci, say to the left of F 1 .
With radius Ai, centres F and F, describe arcs ; and
with radius A'i, centres F' and F, describe arcs intersect-
ing the first arcs in the points marked 1'.
Repeat this construction for the points 2, 3, 4 . . .
obtaining the points marked 2, 3', 4 . . . A fair curve
drawn through the latter points is the hyperbola required.
This construction is based on Theorem 1, Art. 105.
107. Problem. Having given the foci F, F' and trans-
verse axis AA' of a hyperbola, to determine the asymptotes
DE', D'E, and the conjugate axis BB' (Fig. 105).
Bisect A A' in C. With centre Cand radius CF describe
arcs (only half of one shown) intersecting the perpen-
diculars to the axis through A and A' in F, F', >, )'.
Then the lines, of indefinite length, through D', E and
D, F' are the asymptotes.
To obtain the conjugate axis draw a perpendicular
through C, and cut this perpendicular in B, B' by an arc
drawn with A as centre, CF as radius. Then BB' is the
conjugate axis.
Or make CB and CB' each equal to AF.
Note that in the ellipse FB- CA, and in the hyperbola AB-CK
108. Problem. To draw the tangent and normal to a
hyperbola at any point P on the curve (Fig. 105).
Join PF, PF' and bisect the angle FPF' by the line
PT, then PT is the tangent at P.
Draw PG perpendicular to PT, then PG is the normal.
Example. The transverse axis of a hyperbola is 2j" and the
distance between the foci is 2-|". Determine the conjugate
axis, the asymptotes, and draw a portion of the curve.
Draw the tangent at any point and illustrate that this point
bisects that part of the tangent between the asymptotes.
Find the eccentricity and the distance apart of the directrices.
Ans. 1.58", 1.22, 1. 84".
IV
CONIC SECTIONS
107
109. Problem. To construct a hyperbola, having given
the asymptotes XX', YY', and a point P on the curve.
Through the given point P draw lines PAf, PN of
indefinite length, parallel to the asymptotes.
Through the point O, where the asymptotes intersect,
draw a series of radials cutting the lines PM, PN in the
points 1, 1 ; 2, 2 ; 3, 3; 4. 4.
From the points 1 and 1 draw lines parallel to the
asymptotes meeting in I. ; repeat this for the points 2, 2,
obtaining II., and for the points 3, 3, obtaining III., and
so on.
A fair curve through the points I. II. III. ... is the
hyperbola required.
This construction is based on Theorem 3, Art. 105.
Example. Copy Fig. 109 half as large again as shown.
108 PRACTICAL PLANE GEOMETRY chap.
110. Miscellaneous Examples.
*1. CD, CB are conjugate semi-axes of an ellipse. /'J/, is drawn
perpendicular to CD, and PN (on PM or MP produced) is
made equal to CD. Show by construction that if the line MN
slides between the lines CN and CD, the point P on il will
trace out the ellipse. ('889)
Hint. Mark the points N, P, M on tracing-paper ; as this
template moves in the given manner, prick through at /'.
*2. O is the centre of an ellipse, whose major axis lies on AB and
is 3.30" in length. CD is a tangent to the ellipse. Draw the
half-curve above AB. ^895)
Hint. Find the foci by applying Theorem 2, Art. 91.
*3. OH, OK, are tangents to an ellipse at G and E respectively. F
is one of the foci. Find the axes and draw the curve. (1897)
Hint. See Ex. 6, p. 95.
*4. Determine the axes of the ellipse which would touch ab and have
F and F as foci. Without drawing the ellipse find the points
where the line through /' perpendicular to ab would cut it.
*5. Construct the ellipse which has the given points/ 1, and F' as foci,
and which touches the given line ab. ( T 89o)
Hint. Use Theorem 2, Art. 91.
*6. -S" is one focus of an ellipse. Q and B are two points on the
ellipse, and PT is the direction of the major axis. Find the
other focus (geometrically) and draw the ellipse. (1896)
Hint. On BQ prbduced find D such that DQ : DB
= SQ : SB. Draw DZ perpendicular to ST, then DZ is the
directrix. The eccentricity may now be found, and therefore
any number of points on the ellipse. Thus the vertices may
be found.
*7. /"is the focus of a parabola, AB is a tangent to the curve, and
A is on the directrix. Find the axis and directrix, and draw a
sufficient portion of the curve to show that AB is truly tangent
to it. (1893)
Hint. Apply Theorems 2 and 10, Art. 100.
8. Draw two lines, AB, AC, including an angle of 30 . On AC
set off a point P, so that AP=t,". AB is the axis of a para-
bola ; AC is a tangent to the curve at the point /'. Find the
focus and directrix and draw the curve. ( J 894)
Hint. Use Theorems 3 and 5, Art. 100.
*9. The directrix, and two points C and D on a. hyperbola, are
given. If the eccentricity be J, draw the branch of the curve
on which the points lie. (1898)
Hint. A focus is readily found.
IV
CONIC SECTIONS
109
B
Cb/iu She figures double sue
A a
9
4 and 5
o
C
'D
D
E
A
F
7
F'
K
o
F
F
R
Q
CHAPTER V
SPECIAL CURVES
111. Roulettes. Definitions and construction. The
last chapter dealt with the very important curves known as
the conic sections. We have now to consider some other
well-known curves, confining attention to those which have
some application to the work of the engineer and architect.
Taking the curves in the order of their importance from
this point of view, the first perhaps to call for discussion
are those which are generated by the rolling of one curve
on another. The teeth of wheels generally have profiles
thus formed.
.Definitions. If two curves roll upon one another without
sliding, any point connected with one traces upon the
plane of the other a curve which is called a roulette. One
curve is generally fixed or regarded as fixed, and is known
as the base or fixed curve or directing curve ; the other,
which then rolls over the base, is called the generating
curve or the rolling curve.
General method of construction} Let A A be the rolling
or generating curve carrying the tracing-point P, and let
BB be the fixed curve or base.
By the use of a French curve, or otherwise, draw the
curve BB in ink on the paper ; draw the other curve A A
1 See paper by Mr. W. I. Last, on the " Setting out of Wheel Teeth,"
Mm. Proc. Inst. C.E. vol. Ixxxix. 1887, p. 335.
chap, v SPECIAL CURVES in
in ink on stout tracing-paper, or with a needle-point on
thin transparent celluloid, and mark any point P on the
tracing-paper or transparent template. Now adjust the
template until the curves touch each other, and prick off
the point P. Next insert the pricker at the point N
where the curves appear to touch, and rotate the template
through a small angle into a new position, and again prick
through at P. Then insert the pricker at the new point
of contact N, rotate slightly, and prick off P. The locus
of P is the roulette PL, and the above process is con-
tinued until any desired portion of this curve is obtained.
The procedure may be varied by drawing BB on the
transparent template, and AA and P on the paper. Then,
operating as before, the roulette will appear on the moving
transparency. This example affords a good illustration of
relative motion.
The method just described does not ensure absolutely
pure roiling of the curves A A and BB on one another ;
but by taking the steps sufficiently small, the errors come
well within the limits specified in Art. 2, and are not
measurable. Practically, the result is found to be very
perfect, and the roulette is obtained with a quickness and
precision far exceeding that given by any ordinary geo-
metrical construction.
Note. If the template be long and of tracing-paper, a strip or
drawing-paper glued to it will prevent change of shape by buckling.
H2 PRACTICAL PLANE GEOMETRY criAP.
112. Problem. To draw the normal at any point P
of a roulette, and to find the centre of curvature.
The normal and centre of curvature have been denned
in Art. 90 in reference to any curve.
In the present case let the transparency be placed
with the tracing-point coinciding with the given point P
on the roulette, and let this coincidence be maintained
with the pricker inserted at P, while the template is
turned until the curves A A, BB come into contact at N.
Then PN is the normal at P.
For at the instant that the roulette at P is being described the
pricker is at N, and the template is then turning about this point. N
is called the instantaneous centre Of rotation. The tracing-
point at this instant is thus moving at right angles to PN.
The centre of curvature must be on the normal PN.
To find its position, let CD be the common normal to the
curves in contact at N. Let C be the centre of curvature
for the point N on AA, and D that for N on BB. Join
PC, and through N draw a line perpendicular to PN.
Let these lines intersect at Q. Join QD to intersect the
normal PN m O. Then O is the centre of curvature for
the point P on the roulette PL.
113. Problem. (a) To find the length of the given
circular arc AB. (b) To set off a circular arc AB equal to
the given line AE. (c) To mark off a circular arc AD
equal to the given circular arc AB.
The line AE must towh both the arcs at A.
(a) Divide the arc AB into four equal parts at r, 2, 3.
Set off AR on the tangent equal to the chord A\. With
centre R, radius RB, describe the arc BE. Then AE is
equal in length to the arc AB nearly.
(b) Make AR equal to \ AE. With centre R, radius
RE, describe the arc EB. Then AB is the arc required.
(c) Find R as in (a). With centre R, radius RB,
draw arc BD. Then arc AD = arc AB = AE nearly.
Note. These constructions are only approximate. They should not
be used for arcs which subtend angles greater than 90 .
SPECIAL CURVES
"3
Examples. 1. Draw a. fine line in ink on drawing-paper, and a
circle about 2" diameter with a fine ink line on stout tracing-
paper. Roll the circle exactly once round on the line,
operating with the pricker. Measure carefully the lengths of
the circumference thus obtained, and the diameter of the circle,
and calculate the ratio of the two. Ans. 3.14.
2. Repeat this, but draw the line on the tracing-paper, and the
circle on tbi drawing-paper. Ans. 3.14.
3. Draw a circle 3" diameter. Find the length of \ of its circum-
ference by Prob. 1 13. Measure this length and that of the
diameter, and calculate the ratio of the two. Ans. . 7^5-
4. Draw a circle of 3" radius. By Prob. 113 set off on it an arc
equal to the radius, and measure the angle subtended by the
arc at the centre of the circle. Ans. 57.3.
114 PRACTICAL PLANE GEOMETRY chap.
114. Cycloidal Curves. When the two curves which roll
on one another are circles, the roulettes form a class known
as cycloidal curves.
If the tracing-point be on the circumference, we have :
a cycloid when the circle rolls on a straight line ; an epi-
cycloid when it rolls on another circle externally ; and an
hypocycloid when it rolls internally.
If the tracing -point be not on the circumference,
then, when the circle rolls on a straight line, we obtain a
prolate cycloid when the point is inside, and a curtate cycloid
when the point is outside the rolling circle. These two
varieties are also called trochoids. Thus epi- or hypo-trochoids
result when the circle rolls outside or inside a fixed circle.
We shall now give some of the usual geometrical con-
structions for setting out these curves. It should be
understood, however, that they are distinctly inferior, as
regards expedition and accuracy, to the general method
explained in Art. in, in which a transparent template
is used.
115. Problem. To describe a cycloid, having given the
rolling circle.
Let the circle to the left of the figure be the rolling
circle in its initial position, the tracing-point P being then
at o.
Divide its circumference into a number of equal parts,
say twelve, and draw a tangent to the circle at its lowest
point. By Prob. 113 make 02' equal to the circular arc
02, and step off 02' six times from o to o'. Bisect these
lengths, thus obtaining twelve equal divisions, each equal
to one of the twelve equal arcs round the circle.
As the circle rolls to the right, the points of division
on its circumference occupy, in turn, the positions indi-
cated by the corresponding points of division on the
tangent 00'. Also the tracing-point P wi'l ascend and
will cross, in turn, the horizontal dotted lines drawn
through 1, 2, 3, . . . 6, afterwards descending.
SPECIAL CURVES
ii5
Ok'
For example, when rolling takes place, the point 2
descends to 2', and at the same time the tracing-point J }
ascends to the level indicated by the dotted line through 2.
But 2'P. Z - chord 02, hence the construction is as
follows :
Take the points o, 1', 2', . . . 6', in turn, as centres,
and describe arcs to intersect the horizontal lines through
the corresponding points on the circle, the radii being the
lengths of the chords from o to the corresponding points
on the circle.
The second half of the curve may be readily obtained
from the first half from considerations of symmetry. The
two halves are symmetrical with respect to the line -P G 6'.
116. Examples on Problems 114 to 119.
1. Construct a cycloid, the diameter of the rolling circle being 2.5".
Select any point P on the curve and draw the normal at P and
find the centre of curvature.
2- Construct the epicycloid and the hypocycloid, the diameters of
the fixed and rolling circles being 4^" and 1-^" respectively.
Select a point on each curve, for which determine the normal
and centre of curvature.
3- Two circles of 4" and 2" diameters have internal contact. By
the method of Art. in set out the curves traced by a point in
the circumference of one on the plane of the other. Also con-
firm the fact that any point in the plane of the 2" circle will
trace an ellipse on the plane of the 4" one.
n6 PRACTICAL PLANE GEOMETRY chap.
117. Problem. To construct an epicycloid, having
given the radii of the rolling and fixed circles.
Let C x be the centre of the fixed circle, and C that of
the rolling circle in its initial position, the point of contact
o being the initial position of the tracing-point P.
The construction given for the cycloid, Prob. 1 1 5,
applies to this problem with the exception that in place of
the straight lines drawn through the points of division on
the circumference of the rolling circle, there will be a
series of circular arcs concentric with the fixed circle.
118. Problem. To construct a hypocycloid, having
given the radii of the rolling and fixed circles (Fig. 117).
Let C x be the centre of the fixed circle, and C. 2 that of
the rolling circle in its initial position.
The construction, being identical with that of the last
problem, except that the rolling circle is now inside the
fixed circle, is not shown, but the curve is drawn.
119. Problem. To determine the tangent, normal,
and centre of curvature for any point P on a cycloid^
epicycloid, or hypocycloid (Figs. 115 and 117).
Art. 112 should be read again.
First determine the position of the rolling circle when
the tracing - point is at P. To do this, with centre P,
radius C n 6, describe an arc to intersect the locus of the
centre of the rolling circle in C. With C as centre, draw
the rolling circle through P, and find N, its point of con-
tact with the fixed line or circle. Join PN, and draw
PT perpendicular to PN. Then PN is the normal and
PT the tangent at P.
To determine O, the centre of curvature, draw from P
the diameter PCQ.
For the cycloid draw through Q a line perpendicular
to the fixed line to intersect the normal PN produced in O.
For the other curves join Q to the fixed centre C x to
itnersect the normal in O.
SPECIAL CURVES
117
120. Special cases of cycloidal curves. There are two
special cases with which the student should be acquainted.
In the first (a) the hypocycloid is a straight line, being a
diameter of the fixed circle. This occurs when the diameter
of the rolling circle is half that of the fixed circle.
In the second case (b) the circles have internal contact,
but the rolling circle is larger than the fixed circle. The
resulting curve is an epicycloid, identical with that which
would have been described by a circle of diameter equal
to the difference of the two diameters, rolling outside the
fixed circle, as shown at C .
n8 PRACTICAL PLANE GEOMETRY chap.
121. Peculiarities exhibited by curves. We shall now
illustrate some special features which the student may find
in the curves which he has occasion to draw, and give
the names by which such are known to mathematicians.
(a) Asymptotes. A straight line is said to be asymptotic
to a curve when, if the two recede to an infinite distance,
they get nearer and nearer together without limit, but never
actually coincide. Similarly two curves are asymptotic
when they approach nearer and nearer together without
limit, but never actually coincide, as the lengths of the
curves increase without limit. Thus a spiral may be asymp-
totic to a circle, see the figure.
(b) Nodes. If two branches of a curve cross one
another, the point where they cross is called a node. Thus
in the figure the two branches cross at the node and unite
to form the loop. There are two tangents to the curve at
the node, one to each branch. A node is otherwise known
as a double point. If three branches intersect at a point, we
have a triple point, and so on. See the figure.
(c) Crisps. If a point when tracing a curve come to a
place where it stops for an instant and then returns on
itself, so as to generate two branches which have a
common tangent at the point, we have a cusp. Two
varieties of cusps are shown in the figure.
(d) Points of inflexion. If a curve cross its tangent at
any point, the latter is called a point of inflexion. Several
other alternative and equivalent definitions might be given.
Thus a point of inflexion is where three consecutive points
on the curve are in a straight line, or where the radius of
curvature is infinite, the centre of curvature crossing over
from one side of the curve to the other, or where the curve
changes from the concave to the convex on the same side,
or vice versa. Or where the tangent accompanying a point
which traces the curve has its direction of rotation reversed,
being stationary for an instant at the point of inflexion.
SPECIAL CURVES
119
(C)
(d)
120 PRACTICAL PLANE GEOMETRY chap.
122. Envelopes. If a curve move in any definite manner^
there is a certain curve which it always touches ; this is
called the envelope of the moving curve.
Thus let successive adjacent positions of the moving
curve be those marked i, 2, 3, 4 . . . Then a fair curve
drawn to touch these is the envelope of the moving curve.
Let the curves I and 2 intersect in a; 2 and 3 in b ; 3 and 4 in c,
and so on. Then the envelope may also be defined as the curve
through the intersections a, l>, c . . . of consecutive curves, when
these are taken very very near to each other, or when the curve is
moved by indefinitely small steps.
The reader may acquire a very clear notion as to the nature of an
envelope by proceeding thus : Take a piece of transparent sheet
celluloid, and shape one edge of it to the form of the moving curve.
This is most readily effected by tracing the curve on the celluloid by
means of a French curve and needle-point, then bending and breaking
the celluloid along the scratched line. Now let the manner in which
the curved edge of the celluloid shall move be decided in the following
way : On the celluloid draw some curve, say a circular arc, and on
the drawing-paper draw another curve, say another circular arc. Make
the first arc roll on the second as explained in Art. m, but instead of
pricking through at /"use the curved edge of the celluloid as a template,
and by means of it draw a curve on the drawing-paper for each suc-
cessive position.
One method of setting out wheel teeth is by envelopes.
Example. Find the envelope of a straight line which moves so
as to have two points P and Q in the line, 3" apart, always on
two lines pp and qq, which cross one another at 6o.
123. Parallel curves. If a series of normals of equal
lengths be drawn from consecutive points on any curve,
the curve through their ends is said to be parallel to the
first curve, or the two curves are parallel to one another ;
they are equidistant at all points when measured normally.
Otherwise, if a circle move with its centre on any curve,
the envelope of the circle is a curve parallel to the first
curve. This second definition indicates the practical
method usually adopted when drawing parallel curves ; in
applying it, the student will see that the circle has two
envelopes, thus giving two parallel curves on each side of
the original curve.
SPECIAL CURVES
121
% 5 Ar b 6
A curve parallel to a circle is a concentric circle, and
that parallel to a straight line is a second straight line. In
all other cases the parallel curve is of a different character -from
the original one. Thus the parallel curve to an ellipse is
not an ellipse. This is seen very clearly in the figure.
Here the middle curve, an ellipse, is first drawn. Then a
series of circles are drawn of constant radius, with their
centres on the ellipse, and sufficiently near together to
enable the envelope to be drawn freehand. The external
envelope is not an ellipse, though it might appear so to an
unpractised eye. But no one could mistake the inner
envelope for an ellipse, with its four cusps and two nodes.
Note. A cusp occurs on the envelope for a point on the ellipse
where the radius of curvature is equal to the radius of the moving
circle. The outer envelope never has cusps ; the inner envelope only
has cusps when the radius of the circle lies between the greatest and
least radii of curvature of the ellipse which occur at the ends of the
minor and major axes. If A', r be their lengths, a and b the semi-
axes, then r, b, a, A are in geometrical progression. Hence given a
and b, we can find A and r.
122 PRACTICAL PLANE GEOMETRY chap.
124. The involute and evolute of a plane curve.
Definition \. If a straight line roll on a curve, the locus
of any point on the line is called an involute of the curve.
The involute is thus a special case of a roulette, and can be set out
in the manner explained in Art. III.
Definition 2. The locus of the centre of cmvature of any
curve is called the evolute of the curve.
Thus let P v /% /g ... be consecutive points on a curve ; P l O v
P. 2 2 , P 3 O s . . .' the normals ; and O v O.,, 3 ... the several
centres of curvature ; then the fair curve through O v 0%, <9 3 . . . is
the evolute of the curve P v P 2 , P 3 . . .
It will readily be seen that corresponding to a given curve there is
only one evolute, but an infinite number of involutes.
It is shown in works on pure mathematics that any
normal PO to the curve at T is a tangent to the evolute at
O. Also that the difference between any two radii of
curvature is equal to the corresponding arc of the evolute ;
i.e., P % Oi-P x O x = *rc 0,0,; or P 3 3 - P x O x = arc 0,O z ,
and so on. Thus we have the following theorems :
Theorem 1. The evolute of a curve is the envelope of the
normals to the curve.
Theorem 2. If the evolute 00 of a curve PP be drawn,
then PP 'is one of the involutes of 00, and might be traced
by a point P in a straight line which rolls on 00.
Theorem 3. Pi a roulette which is traced by a point P in a
straight line which rolls on any curve, the point of contact O is
not only the instantaneous centre of rotation, but also the centre
of curvature for P.
125. Problem. To draw an involute of a given circle,
and to find the tangent, normal, and centre of curvature
for any point P on the curve.
Divide the circumference into a number of equal parts,
say eight. At the division o draw a tangent, and by the
aid of Prob. 1 13 set off 02' equal to one-quarter the circum-
ference. Step off 02' along the tangent, and subdivide
each of the steps as shown, thus obtaining divisions on the
tangent equal to the arcs of the circle.
SPECIAL CURVES
12-
Draw tangents to the circle at the points, and along
each tangent set off the distance of the corresponding point
on o8' from o. For example, make 44 1 = 04'.
The curve may extend indefinitely outwards.
Let P be any point on the involute. From P draw a
tangent PN to the circle, and find N its point of contact.
Then PN is the normal at P, N is the centre of
curvature, and PT, perpendicular to PN, is the tangent.
This construction is inferior to the method described in Art.
1 1 1.
124 PRACTICAL PLANE GEOMETRY chap.
126. Spiral curves. Let a straight line, starting from a
position OX, rotate continuously in one direction about
O, and at the same time let a tracing- point P move
continuously along the line always receding from or
approaching towards O ; then the curve generated by P is
called a spiral. The point O is called the pole ; OX is the
initial line; OP the radius vector (or radius), and the
angle XOP is called the vectorial angle, for the point P on
the curve.
If during the tracing of the curve the revolving line
make one, two, or more rotations, the spiral is said to
consist of one, two, or more convolutions.
The nature of the curve depends on the law connecting
the motions of P along the line and the line itself round
O ; there is evidently an infinite variety of form.
We shall give the construction for two of the best known
spirals, the laws for which are simple.
In the first, equal amounts of increase in the vectorial
angle and radius vector accompany one another, or, if the
vectorial angles are in arithmetical progression, so are also
the radius vectors. This is the Spiral of Archimedes.
In the second, if the vectorial angles increase by equal
amounts, that is, form a series in arithmetical progression,
the radius vectors form a series in geometrical progression, or,
the ratio of any two radius vectors differing by the same
angle is constant. This is the Logarithmic Spiral.
The involute of a circle is a spiral curve.
127. Problem. To draw an Archimedian spiral of two
convolutions, having given the pole 0, and the longest and
shortest radii OA, OB.
Take OABX as the initial line.
From O draw a series of lines at equal angles with one
another, say 30 , for which angle the lines can all be
drawn with the 6o and 30 set-square, and there will be
twelve radiating lines altogether.
Bisect AB in C, and divide AC and CB each into
SPECIAL CURVES
125
2 78 24- X
twelve equal parts. Then AC or CB is equal to the
increase in the radius vector per revolution, and each of the
divisions is equal to the increase for 30.
Therefore make 01. = Oi, Oil. = O2, Olll. = 0$,
and so on. Then a fair curve through A, I. II. III. . . .
B is the Archimedian spiral required.
Examples on Problems 126 to 129.
1. Draw two convolutions of an Archimedian spiral, the radius
vector increasing from .5" to 2.5" in the two turns.
2. Draw a logarithmic spiral of two convolutions, the least radius
being j", and the ratio of two radii at an angular interval of
22^ being 1.08.
3. Set out a logarithmic curve, taking 16 equidistant ordinates f"
apart ; the least ordinate being J", and the ratio of any two
consecutive ordinates 9:10.
126 PRACTICAL PLANE GEOMETRY chap.
128. Problem. To construct a logarithmic spiral
having given the ratio of the lengths of any two radii
and the angle between them. Let the ratio be 9 : 10
and the included angle 30.
From any pole O draw a series of radials making 30
with one another. This is best done with the 30 and
6o set-square and tee-square.
On any line (preferably one drawn with the tee-square)
mark off oa /line units long on any convenient scale. At
a erect a perpendicular to oa. With centre 0, radius ten
units (on the same scale), draw an arc intersecting the
perpendicular in b. Join ob and produce this line.
Now draw in succession : be perpendicular to ob ; ed
perpendicular to oc . . . ; and ax perpendicular to ox ;
xy perpendicular to oy . . . We thus obtain a series of
lines . . . oz, oy, ox, oa, ob, oc, od, . . . whose lengths are
in geometrical progression, the ratio of any two consecutive
terms being 9:10.
Let these lengths be set off in succession along the
radii from the pole O, i.e. make OZ= oz, OY oy, OX=ox,
OA = oa, OB = ob, and so on.
The fair curve through . . . ZYXABC ... is the
logarithmic spiral required.
One property of this spiral is that the tangent P'P at any point P
makes a constant angle with the radius vector OP wherever P may be ;
the curve is therefore also known as the equiangular spiral.
129. Problem. To draw a logarithmic curve.
Draw any straight line OX, and along it, by applying
the scale, or otherwise, mark off a series of points,
1, 2, 3, 4, . . . at equal distances apart, and erect per-
pendiculars at the points. Set off on the perpendiculars
in succession a series of lengths which are in geometrical
progression, found as in the last problem.
The fair curve through the ends of these perpendiculars
is the logarithmic curve required.
The base XO is an asymptote to the curve.
SPECIAL CURVES
127
/ Z 3 4- 5 6 7 8 9 10 77 7Z
128 PRACTICAL PLANE GEOMETRY chap.
130. Problem. To draw a curve of sines of given
amplitude.
With centre O, and radius equal to the given amplitude,
describe a circle. Draw OF and take any two points D,
E on this line.
Divide the circle and DE into any and the same
number of equal parts, say 16, numbering the parts in
each case from o to 16.
At any point, say 3, on DE erect a perpendicular to
meet a line drawn parallel to OF from 3 on the circle ;
this gives one point on the required curve ; repeat this
construction for each of the remaining points, and finally
draw the fair curve through the points on the perpen-
diculars. This is the curve required.
A point of inflexion occurs where the curve cuts DF.
This curve is the same as the projection of a helrx or
screw thread. See Prob. 369.
The ordinate 33 of the curve is proportional to the
sine of the angle AO3, since the sine is equal to the
ordinate t>2> divided by the radius of the circle. See Art.
13. Hence the name of sine curve or curve of sines.
131. Simple harmonic motion. Definitions.
Suppose a point P to move at a constant speed in a
circular path ; and let PM be a perpendicular from P
to a fixed diameter AB. If this perpendicular move with
P, its foot M is said to execute a simple harmonic motion or
simp/e vibration. The circle in which Amoves is called the
directing circle, and P v~> the directing point.
The amplitude of the vibration is equal to CA or CP;
that is, to half the travel of the vibrating point M, or to
the radius of the directing circle.
The period of the vibration is the time which elapses
while P goes once round the directing circle ; or it is the
time occupied in a complete vibration of M, there and
back.
The phase of the vibration for any position M is the
SPECIAL CURVES
129
E/ F
fraction of the period which has elapsed since the moving
point last passed through its middle position in the positive
direction. Thus in the figure let from A to B be the
positive direction, and let the rotation of CP be clockwise,
as indicated. Then the phase for the position of J/shown in
the figure is the angle DCP, expressed as a fraction of one
revolution. Thus to obtain the phase we might measure
the angle DCP in degrees and divide by 360 . For a
phase of |th, DCP is 45. If M were at A, the phase at
that instant would be f .
Examples. 1. Set out a curve of sines the amplitude OA being
ii" and the base DE 6" long.
2. Measure the series of equidistant ordinates or perpendiculars for
that portion of the curve situated above the base DE, Ex. I.
Calculate the mean ordinate by Simpson's second rule, Prob.
43. Ans. .95".
K
130 PRACTICAL PLANE GEOMETRY chap.
132. Component and resultant motions.
We now consider the path of a point which has two or
more motions simultaneously given to it. The motion which
the point actually has is called the resultant, while the inde-
pendent motions giving rise to this are named components.
In order to fix our ideas, let us suppose that a point
moves uniformly towards O over a length PQ of the bar
AO, while the bar rotates uniformly about a fixed point O
through the angle AOA' ; then the point will arrive at Q
by way of a certain curve PBQ. The point may, however,
be supposed to move from the position P to that of Q' in
either of the two following ways amongst others :
First, let the point move from P to Q while the bar
remains in the position AO ; then allow the bar to rotate
through the angle AOA' ; thus the point arrives at Q.
Secondly, let the bar move from AO through the angle
AOA' without allowing the point to move along the bar,
P will move to P' ; then let the point move on the bar
from P' to Q'. The point again reaches Q'.
In each of these two latter ways the point receives its
component motions in succession. Although the paths are
different, the final position Q' is the same in all three ways.
Now let us further suppose that, in addition to the two component
motions already referred to, the point receives a third component
motion, due to O having a uniform motion from O to O' along 00',
the three taking place simultaneously ; in this manner the point will
arrive at Q" by way of a certain curve PB'Q".
The point may, however, be brought from the position r to that
of Q" by allowing it to receive the first two component displacements
in either of the above two ways, and then allowing O to move along
00' to O', the bar moving from OA' to O'A" without rotation ; thus
the point will arrive at Q". By giving the component displacements
in succession in every possible order it is seen that there are six ways
in which the point may arrive at Q". From these observations the
truth of the following statement will be manifest :
If a point has a number of simultaneous component motions
impressed upon it, it can be brought from any o?ie of its positions
to any other, by giving to it its corresponding component displace-
ments in succession, and in any order.
SPECIAL CURVES
13 1
Q
\ ^""^ia. /
\ ?P?
\B / F
\ /
/
/
1 /
A'
'
^C
/
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/
/
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132
/
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tf'
^ /' 2' J' <' J' ' 7' ^>
133. Problem. A point P moves backwards and for-
wards at a constant speed between two points A and B
in a straight line, and the line has a similar motion
between given limiting positions ab and a'b'. The time
of the first oscillation is double that of the second, and
the point starts from the position a. Determine the path
traced by the point.
Divide ad into a number of equal parts, say four, and
divide ab into double the number of equal parts, that is
eight Draw the dotted lines as shown.
From the conditions it is evident that when AB occupies
the. position n say, the point /"will be in the line iV, so
that P x will be the actual position of P at this instant.
Proceeding in this way, the zigzag path a^'bxa is
obtained, as required.
i 3 2 PRACTICAL PLANE GEOMETRY chap.
131 Problem. A point M has two component simple
harmonic motions of the same period in directions at right
angles to one another ; to trace the curve described by the
point, having given the amplitudes, and the phases at any-
instant.
Refer to Art. 131 for definitions.
Draw any two perpendicular lines intersecting in T, in
which take points Cand C. It is convenient to make TC
and TC each equal to the sum of the given amplitudes.
With centres C and C describe circles with radii equal
to the given amplitudes, and draw the diameters AB, A'B'
perpendicular to TC, TC as shown.
Choosing as positive directions those indicated by the
arrows on the diagram, viz. the directions from A to B and
A' to B', with clockwise rotations, then D and D' are the
positions of the directing points which correspond with zero
phase. Now set off the angles DCB, D'C'P', clockwise,
fractions of one revolution corresponding to the given
phases. In the figure, the angle D'C'P' is for a phase of
T Lth, and is made equal to T V of 360 or 36 . The
angle DCP is 105 , the phase being ifg- or J T .
Next divide the two circles into the same number of
equal parts, say 1 2, P and P being the zero points, and the
points of division o, 1, 2 . . ., o', 1', 2' . . ., proceeding clock-
wise. From corresponding points draw lines respectively
parallel to CT and C'T; their intersections will give points
on the required curve. A pair of these, those from the
points 3, 3', are shown intersecting in Af.,, the others being
omitted to avoid confusing the figure.
The curve is an ellipse, and the circumscribing rectangle
should be drawn as shown, the points where the ellipse
touches the rectangle being determined. Thus to find the
point of contact M. When the tracing-point is at M, P is
at A, between 5 and 6, and P is at m between 5' and 6'.
Therefore find m by setting off the angle 5' Cm equal to the
angle 5CA, then from m draw a line parallel to C'T to
intersect the side of the circumscribing rectangle in M.
SPECIAL CURVES
133
Examples. 1. A point M has two component simple harmonic
motions of the same period and equal amplitudes of \V' , in
directions at right angles to one another ; set out to scale the
curves traced by M, the initial phases being (a) o and o ; (//)
o and \ ; (c) o and y^.
Atis. (a) a straight line ; (/') a circle ; (c) an ellipse.
2. A point M has two component simple harmonic motions in
perpendicular directions. The amplitudes are i" and i^" ;
periods I to 2 ; and initia phases (a) o and O ; (fi) o and j ;
(() o and 1. Set out the paths of M.
Ans. (a) Double-looped curve having a node and a point
of inflection on each branch at the centre; (/>) a para-
bola.
3. Determine the curves traced by a point I\I which has two
component simple harmonic motions at right angles as follows :
Amplitudes l" and i\" ; periods 2 to 3 ; initial phases
(a) o and O ; {/>) o and ^ ; (<) o and T V
4. Trace the curves described by a point M which has two com-
ponent simple harmonic motions in perpendicular directions of
amplitudes 1" and 1^" ; periods 3 to 4 ; and initial phases (a)
o and o ; (l>) j and \ ; (c) j and o.
134 PRACTICAL PLANE GEOMETRY chap.
135. Problem. A point starting from the position P
moves uniformly round the circumference of the circle,
centre C, while the circle turns about the fixed point in
such a manner that the diameter OA moves through the
angle AOA' of 90 and back again without stopping with
uniform angular velocity ; determine the locus of the point
P.
With centre O draw the arc AA' .
Divide the circumference of the circle, centre C, into a
number of equal parts, say twelve, and the arc AA' into
half the number, six.
The method adopted is that of finding a series of
positions of P, giving a succession of points on the curve,
and we shall illustrate the method by determining the
position of the tracing-point when it has moved over one-
twelfth of the circumference of the circle.
Let the point receive its component motions in succes-
sion, Art. 132. First suppose that it receives its circular
motion about C ; this will carry it from jP to 1.
Next, suppose that the circle turns about O until OA
arrives at 0\ ; during this second motion the point turns
about O, moving from 1 to P v where the angle iOJ\ =
angle AOi.
P Y may be determined from the fact that the triangle
Oi'P 1 is equal in all respects to the triangle OAi.
Proceeding in this way, the series of points may be found,
and a fair curve drawn through them.
The path from P 6 to P is a straight line.
136. Problem. OX is a vertical axis, and OA the
initial position of a rod which turns about in a plane
containing OX until it has described an angle which is
double the angle AOX, the point during the same time
moving along OX to 0'. If both of these motions be uniform,
determine the locus of a point which starts from and
moves at a constant speed along the rod from to A and
back again in the time that moves to 0'.
SPECIAL CURVES
135
136 PRACTICAL PLANE GEOMETRY chai\
The point has three component motions : (1) a motion
along the rod ; (2) an angular motion about O with the
rod ; (3) a motion due to the point O sliding down 00'.
Divide 00' into a number of equal parts, say eight ;
also divide OA into half the number, that is four equal
parts. With O as centre describe the arc A4', and divide
this arc into four equal parts. In each case number the
points of division as shown.
We shall show how to determine the position of the
tracing-point when one-fourth of the total time has elapsed ;
that is, when O has moved to the point 2.
Taking the component displacements in succession :
1. Suppose the bar to have its vertical motion bringing
it to the position 02, where Aa =02.
2. Let the bar then have its angular motion ; that is,
with 2 as centre describe the arc aA. making liA. 1 = A2'.
3. Now let the point move along the bar ; that is, make
2P 2 = O2 . Then P 2 is the desired point.
Repeat this construction for the eight positions.
The component motions might have been given in any
order, but probably that indicated will involve least trouble.
The path of P is shown.
137. Motions under mechanical constraint. All the
curves and in fact the lines of all the geometrical figures
we have yet considered may be regarded as having
been described by points moving under some kind of
constraint. A straight line, for example, by a pencil point
guided by a straight-edge. A circle by a point controlled
by a second point (or axis) embedded in the material of the
paper and drawing-board. In other cases we have had
geometrical conditions imposed on the motion of the
tracing-point, and by means of the constraints of ruler and
compasses we have found a series of isolated points in the
required path, and have completed the curve either by
muscular constraint, as when drawing a fair curve by free-
hand through the points, or by the use of a template, such
as a French curve or bent spline, to guide the pencil.
SPECIAL CURVES 137
We shall now give some problems relating to the motions
of parts of mechanisms or machines. The constraint in
such cases may be described as mechanical. The problems
and methods of solution do not differ essentially from those
preceding ; it is only the form of the data that is new. We
are not required to make the mechanism nor a model of it,
so as to allow the moving point in it to actually trace its
own curve ; nor are we even required to make a drawing
of, or to be acquainted with, the constructional details of
the mechanism.
A mechanical constraint always has its geometrical
counterpart, and the first thing is to realise what this is
and how it is to be represented. In the majority of
cases the mechanical constraints consist only of sliding
and turning pairs of elements, represented respectively by
straight lines and circles. The moving pieces may also be
represented by lines.
We thus set out to scale a line or " skeleton " diagram
of the mechanism, putting in only such lines as are
essential to geometrically represent the moving pieces and
the several constraints. This figure, or a portion of it, is
then drawn for a number of positions of the mechanism as
the latter moves through all of its possible positions, and
the corresponding positions of the point under considera-
tion are noted. A fair curve drawn through the series of
points thus determined gives the required path.
This path is always a closed curve for a machine which
works continuously, repeating its cycle of operations, since
in such a machine no point can go off to infinity. This,
however, would not necessarily be the case if we were
finding the envelope of a moving piece instead of the path
of a point.
It will often happen that the piece during its motion
assumes critical ox limiting positions, the accurate determina-
tion of which is desirable in order to draw the curve to the
best advantage. Illustrations will appear in the particular
problems to which we now proceed.
138 PRACTICAL PLANE GEOMETRY chai\
138. Problem. To find the path of a point in the con-
necting rod of a direct-acting steam engine.
The figure is a skeleton diagram representing the
mechanism in one of its positions. CB is the crank, turn-
ing about C, and constraining the crank pin B to move in
a circle. BA is the connecting rod, with the end B centred
on the crank pin, and the end A compelled to keep to the
straight line AC by means of a slide. P is the point in
the connecting rod whose locus is required. The problem,
stated geometrically, would read thus : -
A given line AB of constant length moves with one end
B always on a given circle, centre C, and the other end on a
straight line directed through C ; find the locus of a point
P in AB.
We must draw the connecting rod AB, and thus locate
P, for a number of positions in the cycle, say twelve
positions. To do this in the best way, divide the crank
pin circle into twelve equal parts, beginning at O (most
quickly effected by 30 and 6o set- square). With these
points, o, 1, 2 . . . 1 1 as centres, and radius equal to BA,
describe arcs cutting the line in which A moves, thus ob-
taining the twelve positions of A which correspond to the
twelve positions of B. The connecting rod AB is shown
in the figure for one of these pairs of positions, viz. 1, 1' ;
and the point P 1 is marked off at the given distance from
one end. In determining the locus of Pihe other positions
of the connecting rod were drawn, but for clearness have
been omitted in the diagram.
In this and similar problems we might with advantage again make
use of a transparent template. Thus draw the circle and the line AC
on the paper ; then mark the points A, P, B on tracing - paper or
celluloid, and adjusting the template in succession to a number of
positions, with A on the line, and B on the circle, in each case prick
off the position of P. It is now hardly worth while to divide the
circle into equal parts.
Examples. 1. In a direct-acting engine the crank is I foot long
and the connecting rod 4 feet ; find the locus of the middle
point of the rod. Scale -|th.
SPECIAL CURVES
139
2_-
j
s?
A __
p^=^=
1 r ^^
c
\ ,5
I 6 '
0'/'
^v^
/7
w"
<f
tf
2. The middle point of a Gooch link of 48" radius is guided along
the centre line of a horizontal engine. The ends A, B of the
link are 20" apart and are connected by " open " rods AP, BQ,
60" long, to two eccentrics CP, CQ each of radius 4" and with
angular advance of 30 . Find the motion of the valve at half
gear, the link-block then moving in a horizontal line DD 5"
below the centre line.
Solution. Draw the horizontal centre line KC. Set off angles
KCP= 120 above KC, and KCQ the same below, and make
CP, CQ each 4". With centre C draw the circle through P
and Q. Divide its circumference into twelve equal parts,
numbered o to II, the divisions o and 4 being at P and Q.
With these points as centres, radius 60", draw twelve arcs in
ink, o to 1 1, extending about a foot above and below K. Draw
a horizontal line DD 5" below A".
On tracing-paper stiffened by a lath draw an arc of 48" radius,
on which mark A, B 20" apart, and the mid point of the arc.
Now place this template with its convex side towards C,
and adjust it so that its lower and upper points A and B, and
its mid point O, lie on the arcs o, 4, and the centre line respec-
tively. Then prick through the point > Q where the curve on
the template crosses DD.
Again adjust the template, O being on KC as before, but A
and B now lying on arcs 1 and 5. Prick through D x on DD.
Repeat this for the arcs 2 and 6, determining D 2 ; for 3 and 7,
obtaining D 3 ; and so on.
Measure the displacements Z> , D v D 2 . . . D n of the
valve from the average position of D. A/is. 2.71", 3.08", 2.64",
i-55", 0.13", -1.34", -2.46", -3.01", -2.83", -1.84",
-0.21", 1. 51".
(From which, by Fourier's analysis, ^=3.07 sin (6+ 57)
+ o. 14 sin (20 + 1 03). d is the crank angle measured from 6'A". )
i 4 o PRACTICAL PLANE GEOMETRY chap.
139. Problem. To set out the complete path of the
guiding point in "Watt's simple parallel motion.
This mechanism consists of two " radius rods " or
" levers " CA, DB, centred at C and D, having their ends
A and B connected by a link AB. The guiding point
P is situated in this link, and in such a position that P
divides the link into two segments having a ratio inversely
as the radius rods. That is, AP ': PB = DB :AC. The'
locus of P is a two-looped curve something like a figure
of 8, and in the neighbourhood of the node or double
point the two branches of the curve are very flat, being
almost straight lines. Hence P serves as a point of attach-
ment for a piece which moves to and fro through a limited
range, and requires to be guided in an approximate straight
line, such as the pencil of a Richard's Indicator, or the top
of the piston rod in a beam engine.
To trace the locus of P. With C and D as centres,
describe circular arcs through A and B respectively. Then
A and B move in these paths.
Next trace the link APB on celluloid with a needle-
point, marking the three points A, P, B by short cross
lines. Move this template, placing A in a new position
A' on the arc through A. Insert the pricker at A', rotate
the template until B comes on the other arc at B', then
prick off the point P at P. Repeat this operation a suffi-
cient number of times to enable the locus of Pio be drawn.
Or operate with compasses in the usual manner.
Limiting positions. Observe that the highest possible
position of A is A , where DB and BA come into one
straight lineZ>i? ^ , and where, therefore, DA = DB + BA.
Similarly its lowest position is A v where DA^DB + BA.
In like manner the limiting positions of B are B\ and B 2 ,
where CB 1 = CB. 2 = CA + AB.
Thus to find "A Q and A x describe arcs with centre D,
radius DB + BA, to intersect the path of A in A and A v
To find B 1 and B 2 describe arcs with centre C, radius CA
+ AB, to intersect" the path of B in B x and B. 2 .
SPECIAL CURVES
iji
>Ao
B'\B }
Note also that A B is normal to the curve at P Q , since
for this position A is the instantaneous centre of rotation
of the link. This statement may be tested by trial with
the template, and its truth will then be quite evident.
The end A of the link, having come to the limit of its
movement at A , is there stationary for an instant before
retracing its path.
Examples. 1. In a simple parallel motion the lengths of the
oscillating rods or levers DB and CA are 3 feet and 4 feet
respectively, and the length of the connecting link AB is
2 feet. The mechanism is so set that when DB and CA are
horizontal, AB is vertical. Eind the position of guiding point
P in the link, and set out the complete locus of P. Scale
1" to I'.
^^=C^-^ = 7 f2 ' = - 857 '
Or, geometrically : Join CD to intersect AB in Q, and make
AP=BQ.
142 PRACTICAL PLANE GEOMETRY chap.
4. In
2. Work Ex. i when the centres C and D are moved 0.35'
nearer together horizontally, so as to give the link a slight
inclination to the vertical when the rods are horizontal, as
in Fig. 139.
Ans. AP= - of 2' = 0.857' as before.
7
Note. The best approximation to a straight line is obtained when
the centres C and D are so adjusted that for its working range
the link AB deviates from the vertical equally on each side.
3. In Ex. 1 suppose the centres C and D are both on the same
side of the link, find the guiding point P in the link for an
approximate straight-line motion, and trace the locus of P.
Ans. Produce BA to P, such that
PA : PB = DB : CA, or PA : PB-PA = DB : CA - DB ;
that is, /M=^t 2 = 5 feet.
1 J
Or, geometrically: For the position in which CA and DB
are parallel, join CD and produce it to intersect pro-
duced AB in Q, then produce BA and make AP=BQ.
[n a drag-link coupling the shafts are 6" apart, the drag link
12" long, and the cranks each 30" long. Find the locus of the
middle point of the link. Scale |th.
Or, stated geometrically : Take two fixed points C and D, 6"
apart, and with these as centres describe circles each of 30"
radius. A line AB of the constant length of 12" moves with
A on the circle C, and B on the circle D. Find the locus of
P which bisects AB. Scale -i-th.
140. Cams. We shall conclude this chapter with an
example of a cam. In previous examples on mechanisms
the problem has been the direct one given the mechanism,
to find the consequent motion. We have now the inverse
problem, generally more difficult given the required
motion, to design the constraining mechanism. When a
machine designer requires a complicated motion, he gener-
ally has recourse to a cam or a combination of cams.
Numerous examples are met . with in textile machinery,
printing machinery, and in many other machines.
One form of cam is shown in the figure. It consists of
a flat plate with an irregular contour ABC, capable of
rotating about an axis O, and so giving a reciprocating
motion to a piece AH, which slides in a fixed guide K.
SPECIAL CURVES
142
H
i
1
:
/r
It is obvious that as the cam rotates, the piece AH
will receive a rectilinear motion, the nature of which will
depend on the shape of the curved edge ABC of the cam,
and on how the latter rotates., whether at a constant or
varying speed.
For example, while the cam turns through an angle
which brings B under the end of the sliding piece, the
latter will move upwards through a distance equal to the
difference of the radii OB and OA.
Hence, supposing the cam to revolve at a uniform
speed, we could (within limits) give any kind of motion to
the reciprocating piece by suitably shaping the edge ABC
of the cam. The motion will be repeated with each revolu-
tion of the cam.
In practice the moving piece would require to have a roller pinned
to its lower end and bearing on the cam, in order to diminish the
friction, and to prevent undue wear.
A spring also is generally required to keep the roller in contact with
the cam when the piece is descending. Neither of these is shown in
the diagram.
144 PRACTICAL PLANE GEOMETRY chap;
141. Problem. It is required that a reciprocating
piece guided by a straight slide shall move at the same
constant speed both ways, the change of motion being
effected without interval. Set out the form of the cam
which rotating uniformly will produce the motion, one
revolution corresponding to one to-and-fro movement of
the piece, and the axis of rotation being in the line of the
slide. You are given the diameter of the roller, and the
greatest and least distances of its centre from the axis of
rotation.
Let CAB be the line of the slide, taken vertical, C the
centre of rotation of the cam, and A and B the extreme
positions of the centre of the given roller, which is pinned
to the lower end of the reciprocating piece.
Divide AB into a number of equal parts, say six.
Divide the half-revolution into the same number of equal
angles ; i.e. draw lines through C at angular intervals of
30 (with the 30 and 6o set-square). On these set off
in succession the lengths Ci', C2', C3' . . ., equal to Ci,
C2, C3 . . . Then if the roller were a mere point the
required shape of the cam, on one side, would be the fair
curve through A, \, 2', 3' . . ., which from Art. 127 is
seen to be an Archimedian spiral.
To allow for the roller we draw a curve parallel to the
spiral, determined as the envelope of a circle of diameter
equal to the roller, moving with its centre on the spiral.
See Art. 123. Some of these circles are shown to the
right of the figure.
The right half of the cam is the same shape as the left
half, the line of the slide being an axis of symmetry. This
cam is known as the heart-shaped cam.
Examples. 1. Design a cam to give the following motion to a
sliding piece : Rise of 3" at a uniform speed during first quarter
revolution ; rest during second quarter ; uniform fall of 3" during
third quarter ; rest during last quarter. Diameter of roller |" ;
least distance of its centre from axis of cam 2".
2. It is required that a point P shall move in a straight line with a
speed which increases uniformly from zero during a vertical rise
V
SPECIAL CURVES
145
/
t
5)
.4
,3
.2
\fr
TV
/j2
/ ^*c ^ ^"~""
* jT^ \
' / ^ \
/ / ^ N
5/ / ^
6
I '"
ll ' '
' -' /
/
' '
1 1 -^ /
1 \s /
t-x \ >
n \ /
^ \ >
* \ 1
* \ /
* \ '
^ \ '
^ V
/ /
*s *"*-
^'''
"'--
"6''
of 3" ; the motion is then to be suddenly changed, and the
point is required to fall 3" at a constant speed, the times of the
rise and fall being equal. Design a cam which, while rotating
uniformly, will each revolution impart this motion to P, the
nearest approach of P to the axis of the cam being 2".
Hint. It is shown in mechanics that when the speed increases
uniformly, the distance from the position of rest is proportional
to the square of the time ; that is in this case, for the rise, pro-
portional to the angle turned through by the cam. Now in a
parabola the abscissa AN is proportional to the square of the
ordinate PN (Theorem 6, Art. 100). Therefore in Fig. 103
take A AT 3" to represent the path of P. Construct the para-
bola AQ, and from the points I., II., III. draw lines parallel
to (Wto meet AN in P v P 2 , P y These are the positions of
P at intervals of 45 during the rise from A to N.
Design a cam which shall give a rise at a uniform speed during
the entire revolution, with an instantaneous drop at the end.
L
I 4 6 ' PRACTICAL PLANE GEOMETRY chap.
142. Miscellaneous Examples.
*1. Four equal rods, ab, be, ed, da, form a frame jointed at the
angular points. The frame is also pivoted at a. The point c
moves on the circumference of the given circle. Draw the
curve traced by the point x on the bar be. ( 1S87)
*2. Points P and Q are constrained to move uniformly along the
given lines ab and cd. While P moves from to a and back
again, Q moves from c to d, and while P moves from to b
and back again, Q moves from d to e. Trace the locus of a
point on (IP produced 2" distant from Q. (1889)
*3. and d are fixed pivots about which the bars oa, o'b can freely
revolve ; ab is a coupling bar connecting the free ends of oa and
ob. Draw the complete locus traced by the centre point of
the bar ab. (1S93)
*4. A circle A, of diameter EF, rolls on the line CD with uniform
motion from left to right, starting from E. Another circle B,
whose diameter is half that of A , rolls inside the circumference
of A, also with uniform motion, but from right to left, tarting
at E when A begins to move. Circle B is in contact with
circle A at F at the same time that F reaches the line CD.
Draw the curve described by the centre of circle B. (1897)
*5. A point P revolves round the circle with centre C on the line
OD, with an uniform motion. The point C moves from C to
D and back from D to C, also with an uniform motion, return-
ing to Cat the same time that /"has completed one revolution.
(a) Draw the path of P. (b) Supposing that at the same
time the line OD moves round the point O uniformly, making
a complete revolution while P revolves once round the moving
point C; draw the path of P. (1895)
*0. Two bars, o^a, o 2 b, are pivoted at o l and o 2 respectively. At
i they pass through a saddle which can travel along o^a at
fths the speed at which it can move along o. 2 b. Trace the
locus of i. What is this curve ? (iSSS)
7. Two lines meet at an angle of 6o. Find the locus of points in
the interior of the angle such that the sum of their distances
from the two given lines is constant, and equal to 2|". (1S97)
SPECIAL CURVES
M7
CHAPTER VI
CO-ORDINATES PLOTTING ON SQUARED PAPER
143. The position of a point in a plane. Co-ordinates
of a point. Our object in this article is to show how the
position of a point in a plane may be defined. We shall
illustrate the case by means of a concrete example.
Suppose a person wishes to note the position of some
simple object such as a boy's marble on the floor of a
room, so that he may make a scale drawing which shall
exhibit this position. After measuring the room he will
require to make ttvo further measurements of the position
of the object. There is considerable choice in the latter,
as appears from what follows.
To fix the ideas, let the room be rectangular, 18 feet
long and 12 feet broad, as set out to scale in the figure.
The position in the room of the object P may be ob-
served in the following four ways amongst many others :
1. By measuring the distances of P from any two
adjacent sides of the room. For example,
PN=t.6\ PM = 4.1'.
2. By measuring the distances of P from any two
corners of the room. Thus
OP=8.6', AP= 11. 2'.
3. By measuring the distances of P from one corner
and from one side of the room. Say
^=11.2', PJV= 7 .6',
chap, vi PLOTTING ON SQUARED TAPER
149
B
C
Jca/i of Feet
s
1 I....I ... I....I I
10
4. By measuring the distance of P from one corner of
the room, and the angle which the line from P
to the corner makes with one of the sides ; e.g.
OP=8.6\ AOP =28.4.
Any one of the above four pairs of measurements com-
pletely defines the position of the point on the floor ; any
one pair being given, the others could be determined by
calculation or construction. The two measurements of
any of these pairs are called the co-ordinates of the point ;
so we may have different systems of co-ordinates.
Rectangular co-ordinates are illustrated by Case 1. This
system is the one most generally useful.
Polar co-ordinates are illustrated by Case 4. This method
of defining position is frequently employed.
In this chapter we shall confine attention to rectangular
co-ordinates. The above illustration does not represent
the most general case. It requires to be amplified so as to
include points outside the room on the floor-level. This is
done by the convention of positive and negative co-ordin-
ates, defined in the next article.
Example. The floor of a room OACB is 20 feet square. A small
object P on the floor is 14.7 feet from the side OB and 10.3
feet from the side OA. Make a plan of the room showing
the position of P to a scale of J" to 1' and measure
(a) The distances of P from and A. A/is. 17.9 ft., 1 1.6 ft.
(6) The angles AOP and OAP. Arts. 35 , 62.75 .
(<) The distances of P from R and C. Ans. 17.6 ft., II. I ft.
150 PRACTICAL PLANE GEOMETRY chap.
144. Rectangular co-ordinates of a point P in a plane.
In the plane draw any two perpendicular lines of
reference XX and YY' intersecting in O. '1 hese lines
are called the axes of co-ordinates, or the co-ordinate axes,
or simply the axes. The point O is called the origin.
Frx>m P draw perpendiculars PA/, PN to the axes.
The lengths of these lines are the rectangular co-ordinates,
or the co-ordinates of the point P referred to the axes. The
horizontal distance NP or OM, measured parallel to OX,
is called the abscissa, or x co-ordinate, and is denoted by x ;
the vertical distance MP or ON, measured parallel to OY, is
the ordinate, or y co-ordinate, and denoted by y.
These co-ordinates serve to define the position of the
point P in the plane. It is convenient to express this
position by writing " the point (x, y)." Thus, suppose the
abscissa PJV or x were 3 units long, and the ordinate PM
or y were 5 units, we might speak of P as the point (3, 5),
being careful always to write the abscissa first.
Suppose it were required to plot the point (3, 5) on
paper, this could be done in three ways :
1. Along OX set off OM 3 units long ; draw a perpen-
dicular from M, and on it mark off MP, 5 units
upwards ; or
2. Along OY set off ON 5 units long; from A^draw a
horizontal line, on which set off NP to the right,
3 units long ; or
3. Along OX and OY set off OM and ON 3 and 5
units long ; through M and N draw lines perpen-
dicular to the axes intersecting in P.
Observe that whichever method of plotting be adopted,
in each case we measure from the axes 3 units to the right
and 5 units upwards, in order to arrive at the position of P.
Suppose now the point were at P 2 , the distances of 3 and
5 units from the axes being the same, but requiring to be
set off to the left and dozvnwards, instead of to the right and
upwards. In writing down the co-ordinates, how could
this case be distinguished from the last ?
VI
PLOTTING ON SQUARED PAPER
'5 1
\-P A (3,-5)
The convention adopted is to prefix the minus sign to
the co-ordinates of P 2 , writing them - 2 and - 3, and
considering each as a negative quantity. We should thus
speak of the point P as the point ( - 3, - 5).
We might plot the point P 2 in any one of the three
ways described for P; but in all cases, in order to arrive
at P 2 , we must mark off 3 units horizontally to the left, and
5 units vertically downwards.
Thus a negative abscissa is measured or set off horizon-
tally to the left from the vertical axis ; and a negative
ordinate is measured or set off vertically downwards from
the horizontal axis.
According to these definitions, the co-ordinates of P l
are seen to be - 3 and 5 ; and the co-ordinates of P 3 are
3 and- 5.
We have thus made complete provision enabling us to
define the position of any point on either side of either
axis, that is, of any point in the plane of the axes.
152 PRACTICAL PLANE GEOMETRY chap.
145. The use of squared paper. In order to facilitate
the plotting of points, and especially of curves determined
as a series of points, paper may be used which is covered
with horizontal and vertical lines at equal intervals, ruled
by machinery, and known as squared paper. The lines
serve as horizontal and vertical scales, making ordinary
scales unnecessary, and points are quickly located. For
ordinary work the lines may be one-tenth of an inch apart,
every fifth and tenth line being distinguished by a difference
in width or colour.
A portion of a sheet of such squared paper is here
shown full size; the main divisions are i" apart, and
these (or the half-inch divisions) may be read as units, tens
of units, hundreds. . ., or tenths, hundredths . . ., exactly
as described in Art. 5, with regard to decimal scales.
With care the position of a point may be plotted to within
about the one-hundredth of an inch, if the lines are accur-
ately ruled. Sheets of this paper 18" by n" may be
obtained at a very moderate cost, ruled in fine and faint
blue lines, every fifth line being broader and more con-
spicuous than the rest, and ruled alternately in blue and
red. The red lines are thus one inch apart, as are also the
intermediate broad blue lines.
When using squared paper the axes of co-ordinates may
be chosen so as to have any convenient positions on the
sheet, according to circumstances, and the paper may be
placed with long edges either horizontal or vertical.
Examples. 1. Take a sheet of squared paper, and, selecting the
origin at a corner of a 1" square somewhere near the centre of
the sheet, plot the following points, taking 1 inch as the unit :
(2, 5); (6, 3); (2.5. 4-6); (1.44, 355); (-3. 4);
(-3, -i-5).
Note. After selecting the origin, mark the divisions to the left and
right, and up and down, as for an ordinary scale, but with the
additional negative values.
2. Again selecting the origin near the centre, but taking 1 inch to
represent 10 units, plot the following points :
(25, 36); (-42, 25); (30.4, -40.8).
VI
PLOTTING ON SQUARED PAPER
153
1 1 _
, 1 1 1
1
I I
I I
3. Taking 1 inch to represent 100 units, plot the following points :
(60, 360) ; (55, 400) ; (404, 295).
4. Plot the points (1, 2) and (7, 10) ; join them by a straight line,
and read off the ordinates of those points on the line whose
abscissa are 3, 6, 4.2 respectively ; also read off the abscissa: of
those points whose ordinates are 3, 5.2, 9.6 respectively.
Ans. 4.67, 8.67, 6.27 ; 1.75, 3.4, 6.7.
5. Plot the points (o, 3) and (4, o) ; join them by a straight line,
and find the length of the perpendicular drawn to the line from
the origin. Take j" as the unit. .Ins. 2.4".
Note. The above examples will show the student that considerable
care and thought are necessary in selecting the position of the
origin and choosing the scales, in order to secure the best
results.
154 PRACTICAL PLANE GEOMETRY chap.
146. The equation to a curve. All curves of known
form, such as any of those we have yet considered, have
characteristic properties capable of definite expression in
some way or other. Take for instance the hyperbola.
In the figure a hyperbola is shown in which the asymp-
totes XX, YY' are at right angles, and therefore called a
recta?igular hyperbola. Let P be any point on the curve,
and draw PM, PN parallel to YO, XO. Now take the
asymptotes as axes of co-ordinates, and denote the abscissa
and ordinate of P (PN and PM) by x and y, in the usual
way. It was explained in Art. 105 that one of the properties
of a hyperbola is that the product of PN and PM is the
same for all positions of P on the curve ; or, expressed in the
form of an equation,
PNx PM= constant ;
that is, abscissa x ordinate = constant,
or xy = c . . (1 ).
This way of stating a property which characterises the
curve is called the equation to the hyperbola; or, more
definitely, it is the equation to the hyperbola referred to the
asymptotes as axes.
Thus the equation to a curve is the expression of a
characteristic property in the form of an equation in terms
of some system of co-ordinates and constants.
If the reader be making an acquaintance for the first
time with the idea of an equation to a curve, let him con-
sider well its meaning. Here, in the case before us, x
denotes a distance which is altering as the point moves
along the hyperbola ; so also is y altering ; but their product
does not alter, it is constant so long as we are considering
the same curve.
Whatever curve we may have to deal with, if there be
some geometrical fact which is known to be true for any and
every point on it, then the method of representing the
position of a point by its x and y enables us to express the
same fact by means of an algebraical equation. It is, so to
speak, merely a shorthand way of expressing such fact.
VI
PLOTTING ON SQUARED PAPER
155
Suppose that in the figure PN= 3 and PM= 5, then the
value of c in the equation to the curve will be 3 x 5, or 15.
Accordingly, at the point on the curve where the x is 2A,
the y will be 6 ; at the point where x is 1, y will be 15 ;
if x = - 5, y = - 3, and so on.
Thus the equation to any curve involves the idea of a
point which moves along the curve, and although the
co-ordinates change their values, yet the two co-ordinates
are related to each other in a way which does not alter.
If a different system of co-ordinates be taken, the
equation to the same curve will be different. Suppose we
adopt system 2, Art. 143, and define the position of a point
by its distances from two fixed points. Let the two fixed
points be the foci F and F' of the hyperbola. A well-
known fundamental property of the curve is that the
difference of the focal distances is constant (Art. 105);
i.e. F'F - FP = constant, or r - r = c,
where r and r are now the co-ordinates of P in system 2.
156 PRACTICAL PLANE GEOMETRY chap.
This, therefore, is an equation to the hyperbola. The
corresponding equation to the ellipse is
r + r = c.
Or take system 3 of Art. 143, and let A (Fig. 143) be
the focus, and OB the directrix of the hyperbola. Then if
P be any point in the curve, it is known (see Art. 76) that
AP: PN= a constant number greater than unity.
Or writing r and x for the co-ordinates AP and PN in this
system, the equation is
r = ex ;
where c is a constant greater than unity.
If c be less than unity, the equation represents an
ellipse. If c = 1, the equation represents a parabola.
If the student pursues this interesting and important
branch of mathematics in works on Analytical Geometry, he
will find that a?iy curve which is of definite form has
definite equations which represent it. The form of the
equation depends on the particular system of co-ordinates
which may be adopted as well as on the curve.
It is to be understood that in future we shall keep to
the rectangular system.
Examples. 1. Find the rectangular equation to a circle, radius
a, referred to two perpendicular diameters as axes.
Solution. From any point P on the circle draw perpen-
diculars PM, PN on the two diameters or axes OX,
Y, where is the centre of the circle.
Then, by Euc. I. 47, we have the geometrical property,
PM* + PN 2 =OP*,
which expressed algebraically becomes
2. Using the property of a parabola stated in Theorem 6, Art. 100,
find the equation to the parabola, taking the axis of the curve
as the X axis of co-ordinates, and the tangent at the vertex as
the J' axis.
Solution. See Fig. 100. The property in question is
PN 2 = LL 1 xAN,
or y 2 = /jr,
where / is the latus rectum LL X of the parabola.
vi PLOTTING ON SQUARED PAPER 157
147. To plot a curve, having given its equation. In
the preceding article it was stated that every known
curve had some equation which represented it, such equa-
tion being only a particular way of expressing the law of the
curve.
On the other hand, the converse proposition is true,
namely
Proposition. Every equation connecting the co-ordinates
of a point represents a continuous curve or line of some kind.
We do not attempt to prove this general proposition ;
our object is to illustrate it by showing how to plot the
curve in an actual example where the equation is given.
Let the equation be y o.2t 2 .
A student acquainted with elementary algebra will recognise in this
an indeterminate equation ; that is to say, one that cannot be solved in
the ordinary sense by finding definite values for x and v ; but there
is an indefinite number of solutions, or pairs of values of x and y
which satisfy the equation.
Now the above proposition is equivalent to saying that if all the
points given by the unlimited number of solutions, or pairs of co-ordin-
ates x and r, be plotted, such points will not cover the whole plane
of the paper, but will be confined to definite curves or lines, and will
occupy these lines entirely, without gaps or breaks of continuity. In
fact, that the result is the same as if a point were to move in accord-
ance with the law expressed by the equation, and actually trace the
continuous curve.
Let us now test this by actually plotting a number of the solutions,
and noting the result.
To plot the curve whose equation is_>' = 0.2 x 2 , we take
any values for one of the co-ordinates, and calculate the corre-
sponding values of the other. Thus
Take x = o, then 7 = o.2(o) 2 = o.
Again take x = 1, then y = o. 2(1)" = 0.2.
And take x= 1, thenj = o.2( - i) 2 = 0.2.
X = 2,
y = 0.2(2) = 0.0.
X ' - 2,
y = 0. 2( - 2)' = 0.8.
x = 3
y = 0.2(3)*= 1.8.
*= -3i
y = o. 2 \- 3)= 1.8.
x 4,
y = 3.2, and so on.
i 5 S
PRACTICAL PLANE GEOMETRY
CHAP.
Or we might assign any values to y, and then calculate
the values of x. If this is done it will be convenient to
transform the equation, and write it
o.2.v ? =y, or x 2 = $y ; that is, x= *Jsy.
Now take, say, y - o, then x = V5 x - -
Take_y=i, then # - V5 = 2.24.
Take y = - 1, then a- - ^Z - 5 = impossible.
j/ = 2, x = ^/io = 3.16.
y = 2, x impossible.
*= * v/!5 = =*= 3- 8 7-
x = 4.47, and so on.
J:
3
4>
Before plotting the points, it will be convenient to
arrange the results in tabular form, somewhat as follows :
Table giving Solutions
OF
the Equation j = o.2x 2 .
Values of x
1
2
3
1.8
4
2.24
3.16
3.87
3
4.47
Values of.j'
0.2
0.8
3.2
1
2
4
In plotting these points on squared paper, remember
that the abscissa x is set off horizontally from O V, to the
right if positive, to the left if negative ; and that the
values of the ordinate y are marked off vertically from OX,
upwards if positive, downwards if negative.
The result of plotting the above seventeen points is
seen in the figure.
Note. The subdivisions of the squared paper are omitted in the
figure.
If all possible intermediate pairs of values of x and y
had been calculated and plotted, the result would have been
the curve shown in the figure.
It will be found that the co-ordinates of all points on
the curve satisfy the given equation, Or are solutions ; and
that the co-ordinates of any point off the curve do not
satisfy the equation. That is, the points occupy the curve,
the whole curve, and no place but the curve.
VI
PLOTTING ON SQUARED PAPER
159
The curve does not extend below XX, since negative
values of y lead to impossible values of x. The co-ordinate
axis O Y is an axis of symmetry of the curve. The upper
branches extend to infinity, x and y becoming indefinitely
great together.
The given equation y = 0.2.V 2 , or x 2 = $y, merely ex-
presses the law of the curve that one of the co-ordinates is
proportional to the square of the other. Now referring to
Art. 100, it is seen that this is one of the distinguishing
properties of a parabola ; the curve therefore is a parabola.
On further comparison of the equation with the properties
stated in Art. 100, it is readily inferred that O Y is the axis
of the parabola, XX the tangent at the vertex ; that the
length of the latus rectum is 5 ; and the distance of the
focus from the vertex O is \ the latus rectum ; that is, 1^.
Examples. Plot the following curves on squared paper, after
having first calculated and tabulated the co-ordinates of a
sufficient number of points on each :
1. y 2 = o. 2 x z . 5. x = o.2y 2 .
2. y = o.2(x 3) 2 . 6. y = o.ix 3 .
3. ><-2=o.2(x-3) 2 . 7. X 2 J = 4.
4. y = o.2x' 2 1.2X+11. 8. y 2 = Q.^x 3 -\-2.4.
Note. The first five equations represent the same parabola, but
differently placed as regards the axes. In 7 the co-ordinate axes
are asymptotes to the curve.
i6o
PRACTICAL PLANE GEOMETRY
CHAP.
148. The linear equation Ax + By + C = 0. This equa-
tion is said to be linear, because it can be proved that it
always represents a line which is straiglit. It is also said
to be of the first degree, because it contains no terms, like
x 2 , xy, y 2 , x 3 . . . ; it involves x and y to the first power
only.
The student can easily illustrate by actual plotting, on
squared paper, that the above equation represents a straight
line. Take, for example, the linear equation
2,x - 2y + 6 = o.
Dividing by 2, and transposing for convenience of calcu-
lation, we may write it thus
y-= 1.5* + 3.
Now give to x any convenient series of values, calculate
the corresponding values of y, and tabulate the series of
solutions as described in the last article. Thus
Table giving Solutions of the Equation y =1.5x4-3.
Values of x
n
- 2
- 1
1
2
3
Values of y
-i-5.
i-5
3 4-5
1
6
7-5
The figure shows these points plotted, with the straight
line drawn through the points.
One readily observes from the table or the figure that
as x increases by 1, y increases by 1^; or thatjy increases
one and a half times as fast as x. This fact shows the
locus to be a straight line.
We give one more example. Let the general linear
equation Ax + By + C o be transposed into the form
x y
- + j= i,
a
then a and /> are the distances from O of the points A and
B in which the line intersects the axes (9 A' and O Y. These
distances are called the intercepts of the line. Take, for
instance, the equation previously considered,
VI
PLOTTING ON SQUARED I'AI'ER
161
Y
G
5
4- /
B
3
Z
/
X'
A
X
3 /
?
i
1
z
j
-/
-z
Y'
$x - 2y + 6 = o,
or 3-v - 2_r = - 6.
Dividing throughout by - 6, this becomes
3- v
-6
x
2V
= I,
or
+ ^=i.
- 2 3
Now refer to the figure, in which this line is drawn ; it
will be seen that the intercept a or OA is - 2, and the
intercept b or OB is 3.
Examples. Calculate, tabulate, and plot eight points from eacli
of the following equations. Observe that in each case the
series of points lie on a straight line.
1. x=y.
2. x +y = o.
3. x+y=i.
4. x + r +1=0.
5. y = 2x - 1
6. y- -\x+ 1.
7. y = 2x.
8. y= -%x.
9. x = 2y.
10. x y
3 2
11. -r + y
2 7,
I.
12. 2. 35.1- + 3. 1 7j'- 4. 86 = 0.
M
162 PRACTICAL PLANE GEOMETRY chap.
149. Problem. To plot a straight line, having given
its equation, say 4x + 3y - 12 = 0.
First Method. Select any pair of convenient values of
x (or y) and calculate the corresponding values ol y (or x);
plot the points on squared paper ; draw the straight line
through the two points thus plotted.
Thus in the given equation
Put x = o, then y = 4.
Put x = 4, then 7 = J(i2 - 16) = - 1.33.
In the figure the points (o, 4) and (4,-1.33) are
plotted, and the required line is drawn through them.
Second Method. Transform the equation into the shape
xy . .
-- + -=1; the intercepts a and b are then given without
a b
further calculation.
Thus the given equation may be written
4.V+ &= 12,
x v
or - + - = 1 .
3 4
The intercepts are now seen to be 3 and 4, and these
lengths are marked off from O on the diagram, viz. at
A and B.
150. Problem. Having given a straight line and the
axes of co-ordinates (on squared paper), to determine the
equation to the line.
First Method. Assume any linear equation for the
line, say, y = nix + c, where m and c require to be found.
Read off the values of x and y for any two points on the
given line. Insert these pairs of values of x and y in
succession in the given equation. We thus have two
equations from which to determine /// and c. An illus-
tration will make this clear.
Let the given line be PQ in the figure. Select any two
points on the line, say, P and Q.
Read off the co-ordinates of P, viz. x = o, y =2.7; and
the co-ordinates of Q, x = 7, y 4.4.
VI
PLOTTING. ON SQUARED PAPER
163
Y
At
9-
O
1 -
A
i
2
*X
-1
-2-
149
5
Y
4.
\B
^g,
3
P
2
1
\4
1
2
3
4
jN
$
7
150
Now insert these pairs of values in the assumed equation
y = mx + c, and we get
2.7 = o + c,
4. 4 = ;// x 7 + c.
From which
'=2.7,
AA-c _ 4-4- 2-7
7 7
;//
0-243 ;
and the required equation to the line is
y = 0.243.V + 2.7.
Second Method. If the points where the given line inter-
sects the axes are available, and not too close together,
read off the intercepts a and b, and insert them at once
in the equation,
x y
- + j= 1.
a
Thus, if the given line were AB (Fig. 150), the inter-
cept OA or a is 5.6, and the intercept b or OB is 4.7 ;
therefore the equation to the line AB is
x y
5-6 4-7
1 .
1 64 PRACTICAL PLANE GEOMETRY chap.
151. Examples. 1. Plot the following straight lines on squared
paper. In each case adopt the method of plotting which seems
the most suitable.
() y = 3-* - 4-
(b)y = sx.
(c) J = 2.r- 3 .
(d)y=-^x.
tf) >V = 2.
('<) jl' = 3-
(*) x = o.
(/) j< = o.
H-
(w) 2.7x4-3.97' - 6.2
!-;-
() 3-5 v - i-U'+ 2 -3
2. Measure from your diagrams for Ex. I the co-ordinates of the
points of intersection of the lines c and d ; also of e and_/" ; and
of in and ;/. In each case verify the result by calculation ; that
is, by solving the three pairs of simultaneous equations.
Ain. (1.2, -o.6) ; (4.39, -.488) ; (-0.13,1.69).
3. Determine the equations to the ten lines given in the figure
on the opposite page. Answers
x y ,
1. + = 1. 6. y=i.i2$x.
4.8 9.1
2. y=.$x+5.5 7- J= -.83-r.
3. r = 6.43.1- 23. 1. 8. y \.2.\x 4.6.
4. .7' = 2.9X + 8. 4. 9. J = X-S-2.
5. r = .8x + .8 10. j= -2.33x4-2.
4. Read from the diagram opposite the co-ordinates of the point of
intersection of lines 1 and 2. Confirm the result by solving the
simultaneous equations which are given as the answers to 1 and
2, Ex. 3. Ans. (1.5, 6.25).
5. Measure, from the diagram opposite, the "slope" of the line 6,
that is the tangent of the angle which the line makes with OX.
Find the angle from the table of tangents on p. 20. Am. 6.7
in 6 or 1. 16 in 1 ; 49. 3 .
6. Find the slope or gradient of the steepest portion of the road
shown in Fig. (b), p. 167. Aus. 860 feet in 2 miles ; i.e. 1
in 12.3, or 0.0813 m L or 4-7-
7. After having plotted the lines of Ex. 1, measure the angle (1)
between lines {a) and (6) ; (2) between lines (c) and (d) ; and
(3) between (e) and (/). Ans. (1) o ; (2) 90 ; (3) 90 .
Note. Observe (1) that in is the slope of the line y = mx -f- c ;
(2) that the lines y = nix + c, and y = mx + c f are parallel;
(3) that the lines y = mx + c andj'= x4V are perpendicular.
VI
PLOTTING ON SQUARED PAPER
165
1
10
9\
81
'\ /
7
\
2^
\
3
/ 6
^O
4
\| /
/
1/
J
\ \/y\ i
1 \ / 1
2\
i/ ' \ 1
/A \\
/ 1 \
/
V/^'
-/J j / \/
/
'
-/
f x ~
' /
-/
N
1
1
3 1
4 \
5
6
A,
-9
-3
\-4
\
'9
~ ,'j
5
imiiii
6
\io
r
1 '
X
i66
PRACTICAL PLANE GEOMETRY
CHAT.
152. Plotting the results of observation and experi-
ment. As will be explained in a future chapter, any
quantity capable of numerical measurement may be repre-
sented graphically by a finite straight line set out to a
specified scale. The co-ordinates of a point, being lengths,
may be taken to represent two such quantities. We may
extend this to two quantities of variable magnitude, which
are mutually related in some manner, and thus exhibit the
nature of the relationship by means of a curve. From
many familiar examples we select the following :
(a) Load-elongaiion diagram. In testing the tensile
strength of a specimen of mild steel, one square inch
cross sectional area, if the loads and the corresponding
elongations are noted at intervals during the operation,
and subsequently plotted as ordinates and abscissae re-
spectively, the load-elongation diagram is somewhat like
Fig. (a), its exact shape depending on the nature of the
steel. Many testing machines are arranged so that the
load-elongation diagram may be drawn automatically.
bnspersq.in. (a)
30
\ \
\
^^~ "v
<^ s
L L.
I
i
J/J
1
/ft
elonaati<m
Z"
VI
PLOTTIN T G ON SQUARED TAPER
167
(/>) Contour road- map. A cyclist may wish to know
the elevation above the sea-level, or the gradient at any
point of his route ; this is well given by a contour road-
map, Fig. (/>), in which the abscissas are miles travelled,
and the ordinates are heights in feet above the sea-level.
LffilaLTTtTiTiYi' i il
1500 __
5 miles
10
1500
feet
woo
500
Sea-
Level
(c) Price chart. A manufacturer who studies closely
the variations in the market prices of materials day by day
would be interested in diagrams like that of Fig. (c), which
shows the fluctuations in the price of the metal tin during
the month of June 1898, the abscissae being market-days,
and the ordinates the prices in pounds per ton, as quoted
in the London metal market.
tper
71
to/v
W
70
/
69
/
\
/
\
68
/
\
/
67
1
/narked da/us
i6S PRACTICAL PLANE GEOMETRY chap.
153. Choice of scales. In plotting curves such as those
described in the last article, it is important that the scales
be judiciously chosen. The main points to be attended
to are now described.
The horizontal and vertical scales may be chosen quite
independently of each other, even when both co-ordinates
represent actual lengths, as in the case of the contour road-
map of Fig. (/>), page 167.
The zero points of the scales need not be on the sheet.
Thus in Fig. (c), page 167, the vertical scale begins at ^67
per ton, as it would be useless to show anything lower
than this. Note also that in Fig. 155 the efficiency Scale
begins at .40.
The scales should be so chosen that the curve extends ivell
over the sheet, from bottom to top, and from extreme right
to extreme left, and is thus not dwarfed either way.
If the plotted points lie approximately on a straight line,
the best result is obtained when the line is about equally
inclined to both axes. See the line in Fig. 155. We should
avoid adopting scales which would cause the line to be
inclined at an angle less than say 30 to either axis.
We may point out some of the bad effects -which result from
the neglect of these precautions. At the top of Fig. (/'), page 167,
the contour of the road is drawn in its true horizontal and vertical
proportions ; so drawn it would not show with sufficient distinctness
the variations in height and slope of the road, to be of practical use
to the cyclist. The road would appear to he nearly flat notwith-
standing that for a couple of miles there is a gradient of 1 in 12,
almost too steep to be ridden down. In the lower figure the vertical
heights are magnified twelve times.
Again, in Fig. 155 at the top, we have set out the efficiency-
resistance curve with the vertical scale reduced to y^th, to show that
under these circumstances the curve might easily be mistaken for an
approximate straight line.
Or again, in Fig. (<?), page 166, the Inst portion of the diagram
appears to be a perfectly straight line, nearly vertical ; in order to
know whether the line is actually straight or not, it would be neces-
sary to enlarge the horizontal scale for this portion of the figure.
The enlargement would require to be from fifty to one hundred times,
in order to comply with the condition stated above.
VI
PLOTTING OX SQUARED PAPER
169
J
^
^
V
!
/"
1
/
-3
6
3
6
0"
9
A
V
/
-1"
::
154. Interpolation. Maxima and minima. When we
are given the values of a quantity for a series of intervals,
we can readily find any intermediate value by plotting the
given series. This is called interpolation. VVe can also
determine in a similar manner the maximum or minimum
value of a quantity, where such exists, if we are given a set
of values of the quantity ranging on each side of the
maximum or minimum. The following example should
make this clear :
Example. In a steam engine the displacements of the slide valve
from its mean position, corresponding to a series of angular positions
of the crank, are given in the following table : determine
(a) the displacement of the valve when the crank angle is io ;
(/>) the angle of the crank when the valve displacement is zero ;
(<) the maximum displacement of the valve.
Angle of crank .
-6o
-3
o
3o
6o
2-95"
90
Displacement of valve
-.91"
.80"
2.17"
2-93"
2.23"
In the figure above these positions are shown plotted on squared
paper, with crank angles as abscissae, and valve displacements as
ordinates. & fair curve is drawn freehand through the six points thus
obtained. The required results are then read off from this curve.
Thus we obtain : Ans. (a) 2.49" ; {/>) -43.8 ; (c) 3.06".
170
PRACTICAL PLANE GEOMETRY
CHAP.
155. Laboratory test of a crane. The following numbers
were obtained by testing a small crane under different
loads :
Load raised by
crane, in lbs. or
Resistance, /_
o
1
20 | 40
60
80
100
120
_____
38.1
140
1
Force exerted at
handle, in lbs. or
Effort, E .
4-5
I0.7 1 15.6
1
21.7
27.0
32.8
43.6
Efficiency, / (cal-
culated)
o
i
47 i -64
1
7i
74
,6
79
.80
The gearing of the crane was such lhat, but for friction, the force
required at the handle would have been \ the load raised. The
efficiency is calculated by dividing this force (without friction) by the
actual force. Thus when So lbs. is being raised, the effort required if
there were no friction would be 20 lbs., whereas the actual effort from
the table is 27 lbs. The efficiency for this load is therefore 20-^27
= .74, or 74 per cent.
Now, plotting the efforts and resistances as co-ordinates
on squared paper to suitable scales, we obtain points which
are seen to lie very nearly on a straight line.
The best average position amongst the points for this line is readily
judged by applying to the diagram a piece of black thread, or a piece
of tracing-paper on which a straight line has been ruled.
The efficiency-resistance curve is shown in the same
figure. The points on it are plotted, and a fair curve
drawn so as to lie evenly amongst them, so far as can be
judged by sight. This curve may be used to correct
errors of observation.
The two curves may be said to be set out on a resistance
base ; the horizontal scale of resistance is marked in the
figure. The efforts and efficiencies are plotted as ordinates,
their respective scales being shown on the left and right of
the diagram.
VI
PLOTTING ON SQUARED PAPER
171
100
50
Elbs
O ZQ 40 60 80 100 120 R 140 H>S.
155
]'/2
PRACTICAL PLANE GEOMETRY
THAT.
158. Examples on Plotting.
1. Plot the following values of /, the pressure in pounds per
square inch, and t the temperature Fahrenheit, of dry saturated
steam ; join the points by a fair curve, and read off the value
of/ when t is 293 . Ans. 60.4 lbs. per sq. in.
\
t
212 23O
248
266
284
302
320 ! 338 ; 356 374
392
p
14.7 j 20.b
1
28.8
39- 2
52-5
69.2
J
89.7 1 15. 1 j .45.8 182.4
1 ! 1
225.9
2. The values of A, the average attendance (in thousands) at the
primary evening schools in England for the years 1S87 to 1S97
are given ; obtain a curve showing these values.
Year
1887
1888
1889
1890
1S31
1892
1893
1 '-'94
1895
1S96 1897
A
30.6
33-3
37-i
43-3
52.0
65.6
81. 1
"5-5
129.5
147
179.6
3. Find one root of the cubic equation 2.v 3 - $x - 16 = 0.
Method. Let y=2x 3 - 3-r 16. Give to x the values 1,2,
3, etc., and calculate the corresponding values of_j'. Plot these
values of x and y on squared paper and draw a fair curve
through the points so obtained. Read off the values of x where
y is zero, that is, for which 2x 3 ix - 16 zero.
To obtain a more correct solution, after observing that one
required value of .v lies between 2 and 3, take closer values of
x, say 2.1, 2.2, 2.3, etc., and after calculating the values of y,
plot these new values of x and_y to a bigger scale. Ans. One
solution is .1=2.25.
N.B. In this manner one or more solutions to any equation
may be obtained.
4. The half-ordinates of the load water-plane of a vessel are 12
feet apart, and their lengths are 0.5, 3.8, 7.7, 1 1.5, 14.6,
16.6, 17.8, 18.3, 18.5, 18.4, 18.2, 17.9, 17.2, 15.9, 13.4,
9.2, and 0.5 feet respectively. Determine (1) the total area
of the plane in square feet, (2) the position of the centre of
area of the section of the vessel made by the water-plane.
Ans. (1) 2649 sq. feet : (2) 102.8 ft. from the first ordinate.
Hint. (1) Find the mean ordinate as in Art. 43, and
multiply it by the total length to obtain the area. (2) Multiply
each mid-ordinate by its distance from the first ordinate, and add
these products together ; divide the result by the sum of all
the mid-ordinates.
vi PLOTTING ON SQUARED PAPER 173
157. Approximate linear laws. In Art. 155, referring
to the crane, it was seen that when the force applied to
work the crane, and the load raised that is, the effort and
resistance were plotted as co-ordinates on squared paper,
the points obtained lay very nearly on a straight line ;
the law connecting the two is therefore said to be linear,
or approximately linear. The determination of the laiv of
the crane is the same as that of finding the equation to this
straight line. The line was located by means of a stretched
thread or ruled tracing-paper.
We proceed exactly as in Prob. 150. Let the equation
to the line be
y = ax + b,
or E = aR + />,
since the ordinate y is the effort E, and the abscissa x is
the resistance R.
To find the constants a and /', select any two points on
the line, say P and Q, near the ends. Read off the
co-ordinates of R and Q. Thus
for R, E = 4.gi lbs., R = o ;
and for Q, ^ = 43.8 lbs., R= 140 lbs.
Substitute these values in the above equation, and we
obtain
4.91 lbs. =0 + b,
43.8 lbs. ax 140 lbs. + b ;
from which
b = 4.91 lbs.
43.8 lbs. - 4.91 lbs.
a = - = .278.
140 lbs.
Therefore the equation to the effort-resistance line, or
the law of the crane, is
. = .278^+4.91 lbs.;
Or, the force that must be applied at the handle in
order to raise any load is equal to 4.91 lbs. plus .278
of the load.
174
PRACTICAL PLANE GEOMETRY
chap
158. Miscellaneous Examples.
1. In a Stephenson link motion the following measurements
were made : -
Angle of crank .
-45
-i5
i5
45
75
105'
Displacement of valve
-25"
.98"
1.86"
2.24"
2.05"
1.32"
Determine (a) the displacement of the valve when the crank angle
is zero ; (b) the angle of the crank when the valve displacement
is zero ; and (<') the maximum displacement of the valve.
Ans. (a) 1.47"; (b) -39.2 ; (c) 2.25".
2. The following numbers refer to the test of a crane, determine
the linear law of the crane.
Effort E, lbs.
7
4
36
21
65
28
93
35
4-'
49
56
63
Resistance R, lbs.
122
151
183
2l6
248
Ans. E = o. 23 A'4-6.54 lbs.
3. Try if an equation of the form xy = ax-\-by approximately repre-
sents the relation between x and y, pairs of values of which are
given in the table below, and if so, determine the mean values
of a and b.
,
Values of j'
5
6
28
7
8
9
10
1 1
Values of x
18
54
133
-455
- 1 1 1
-65
Ans. a = 8. 7, b = - 13, or xy = 8.jxy 13.
Hint. The equations xy = a x + by may be written in the form of
1 = a- + b-. Therefore calculate and plot the values of the
y x
reciprocals - and - as co-ordinates. If these are called Fand
y x
X, we have then to determine a and b in the approximate linear
law (if such exists) 1= aY+bX, as in Arts. 150 or 156.
VI
PLOTTING ON SQUARED PAPER
1/5
A series of pressures p and volumes v of saturated steam, as
determined experimentally, are given in the table below. The
logarithms of these quantities are also given. By plotting log/
and log v, try whether a relation of the form log/ + <z log vb
(i.e. pv a = cox\sL) holds approximately between them, and if
this is found to be the case, determine the best average values
for the constants a and /'.
Properties of Saturated Steam
Pressure, /. Lbs.
per sq. inch
Volume, v. Cubic
feet per lb.
I.o6
2.88
122
.462
2.09
6.86
14.7
26.4
1. 17
1.42
28.8
14.0
1.46
52-5
7-97
89.7
4.82
146
3.06
226
2.02
313
53-4
.837
log/ .
.020
2.50
1.72
i-95
2.17
483
2-35
log 7' .
i-73
i-'5
.902
.683
35
Ans. (1=1-0$; 3 = 2.66 (or / 1,05 = 46o).
5. A log of timber 20 feet long has the following cross-section
areas at the given distances from one end. Find the volume
in cubic feet.
Distance from one end in feet
2.6
5
7.8
12
3-5
15
17.6
3-i
20
1
Area in square feet . . ' 5.0
1
4-3
3-8
3-6
3-3
3-
A?is. 72. 7 cub. ft.
Hint. It will be observed that the distances between the given
sections are unequal. In such a case first plot the given
numbers on squared paper, and then divide into, say, 10 strips
of equal width. Find the mean cross-sectional area by the
mid -ordinate method, and multiply by the total length of
the log.
SFXTION II
PRACTICAL SOLID GEOMETRY, OR
DESCRIPTIVE GEOMETRY
CHAPTER VII
POSITION IN SPACE DEFINED AND EXHIBITED
159. Introduction. Hitherto the points, lines, and
figures in the various problems taken have been confined
to one plane, represented by the plane of the paper ; we
now pass on to the more general case and treat of the
geometry of space. Our diagrams must now represent
the three dimensions of length, breadth, and thickness,
whereas previously we were concerned with only two,
length and breadth.
It is necessary to distinguish between pure and practical
solid geometry.
Pure solid geometry deals with the geometrical relations
which exist amongst points, lines, and surfaces in space.
Practical solid geometry shows how to exhibit these rela-
tions by scale drawings which can be measured. We have
an illustration of the former in the eleventh book of Euclid,
where a number of definitions relating to solid figures are
vii POSITION IN SrACE DEFINED AND EXHIBITED 177
given, followed by the propositions, arranged in strict
logical sequence and proved. The diagrams are of
secondary importance, being generally semi -perspective
sketches, just sufficient to convey an idea of the form of
the solid figure. It is necessary that the student be
acquainted with the definitions and some of the proposi-
tions of Euclid XL, and for convenience, these are given
in an appendix at the end of this volume.
In practical solid geometry we are concerned not so
much with the proofs of propositions, as with the methods
whereby the relative positions and forms of figures in space
of three dimensions can be exhibited on a surface such as
a sheet of drawing-paper, which has only two dimensions,
and in such a manner that these positions and forms can
be ascertained from the diagrams or drawings by direct
measurements to scale.
As in the preceding section, we shall endeavour to
enforce upon the student the importance of good draughts-
manship. A leading idea throughout should be that in
making practical computations, graphical processes may
with advantage often be employed in preference to other
methods. One is only properly equipped for the work when
accuracy in execution has become habitual.
The special difficulty which the beginner experiences is
to be able to conceive clearly the connection between the
figures drawn on paper and the actual figures in space
which they represent. This difficulty is greatly lessened
in the initial stages by making free use of models. The
problems of this introductory chapter are arranged so
that the student may easily make his own models for all of
them. It has been found that by their use very clear
and precise notions are formed at the start, and an
excellent method of working is introduced.
The problems relate to ways of exhibiting the positions
of points, lines, and planes in space, and of finding the
distances and angles between them. They form the basis
of the method by which solid form is defined by drawing.
N
178 PRACTICAL SOLID GEOMETRY chap.
160. The position of a- point in space denned by rect-
angular co-ordinates. We have seen in Art. 144 how the
position of a point in a plane may be defined by reference
to two perpendicular axes. In like manner the position of
a point in space may be defined by reference to three
mutually perpendicular planes. For example, the situation
of a small object in a room is known if we know its height
above the floor and its distance from each of two adjacent
walls.
Let the planes of 'reference be as shown in the figure, one
horizontal and the other two vertical, and at right angles to
each other, the latter being distinguished as the front and
side vertical planes. These planes intersect in three lines,
also mutually perpendicular, called the co-ordinate axes, one
being vertical and the other two horizontal. The point
common to the three axes and the three planes is called
the origin, and is denoted by the letter O, the three axes
being labelled respectively OX, OY, OZ, the latter being
the vertical one. The three planes of reference may
also be designated as the planes of X Y, YZ, and . ZX
respectively.
The position of any point A may now be defined by
its perpendicular distances Ad, Ad' , Aa from the three
planes of reference. These three distances are called the
rectangular co-ordinates of the point, and may be denoted by
x, y, z. The co-ordinate x is the distance of A from the
plane of YZ, measured parallel to the axis OX. Similarly
for the other two: y is measured parallel to OY, and z
parallel to OZ. Thus x and y are the horizontal co-
ordinates, and z is the vertical co-ordinate.
A co-ordinate has a negative value when the point is
situated on the other side of the plane from which that
co-ordinate is measured. Thus z would be negative for any
point below the horizontal plane XY. And x would be
negative for any point at the back of the front vertical
plane YZ. Similarly, y would be negative for any point to
the left of the plane XZ.
vii POSITION IN SPACE DEFINED AND EXHIBITED 179
The student will note that the co-ordinate planes, when
extended eacli way from 0, divide the neighbouring space
into eight trihedral angles. He should also note that when
the co-ordinates of a point are given in sign and magnitude,
the signs tell us in which trihedral angle the point is
situated, and the magnitudes give lis the exact situation in
this angle. Thus the position of any point in space is
completely defined, without ambiguity, by its three rect-
angular co-ordinates.
The point whose co-ordinates are x, y, z may for short-
ness be referred to as the point (x, y, z).
Examples. State in which trihedral angles the following points
are situated :
1. The point ( - I, 1, 1).
Ans. Behind YZ, to the right of ZX, above XY, or in the
back-right-upper angle.
2. The point (1, - 1, 1).
Ans. The front-left-upper angle.
3. The point (1, - r, - 1).
Ans. The front-left-lower angle.
4. The point ( - I, -I, - I).
Ans. The back-left-lower angle.
i8o PRACTICAL SOLID GEOMETRY chap.
161. The position of a point in space exhibited by
projection. In descriptive geometry it is not enough to
define the position of a point in space. It is further neces-
sary to set out this position by a scale drawing.
As the drawing is necessarily confined to one plane, and
the figure which the drawing represents is not plane, some
convention is required in order to connect the two, and
so render the signification of the drawing precise and
definite.
To illustrate, let the student take a quarter of an
imperial sheet of drawing-paper, and on it rule two perpen-
dicular lines, distant about 5" from the upper and left
edges, as shown in Fig. 161 (a) at ZY, ZX. Then cut the
paper along the horizontal line from O to Z, and indent ! it with
a blunt instrument from O to Kand Z to X. Now fold the
paper along the indented lines, and secure the overlapping
parts on the left by means of a paper-fastener. A partial
model of the planes of reference, Fig. 1 6 1(/>), is thus obtained,
comprising the front-right-upper trihedral angle, the co-ordi-
nates of points situated in which are all positive.
Suppose now it is required to set out the position of a
point A, whose co-ordinates x, y, z are given.
Unfold the model, and along OX and OY mark off 01
and Om equal to x and y, and along OZ, OZ set off On,
On, each equal to z. Through /, m, and n draw the lines
parallel to the axes, as shown, intersecting respectively in
a, a , a". Then this plane figure, or unfolded model, with
its lines and points, is the scale drawing or diagram which,
properly interpreted, represents the position of the point A
as regards the three planes of reference.
If this figure were given, then in order to determine
from it the x co-ordinate of the point A, we should measure
the length of 01, ma, or na" ; to ascertain the y co-ordinate,
we should measure Om, la, or na' ; and for the z co-ordi-
nate we should measure any one of On, On, ma', &&a".
If it were desired to show exactly what the drawing
represented, we should fold it along the lines of the axes,
vii POSITION IN SPACE DEFINED AND EXHIBITED 181
161(a)
z
f
n
i a'
/
f
\m
z
n
CV"
I
>a
Y
X
161(b)
7^7y
so as to obtain the model of the planes of projection ; and
to show the actual point A in space and the three per-
pendiculars from it, we might use a small sphere with three
projecting pieces of wire, Aa, Aa, An".
The points a, a, and a" are called the three projections
of the point A. But some formal definitions are required,
and we give these in the next article.
1 82 PRACTICAL SOLID GEOMETRY chap.
162. Definitions relating to projection. If from the
various points which may be supposed to constitute any
object, straight lines proceed to meet a plane, the object
is said to be projected on the plane.
The straight lines are called projectors.
Parallel projection is the case in which the projectors
are all parallel to one another.
Radial or perspective projection that in which their direc-
tions all pass through one point.
Parallel projection is subdivided .into oblique and
orthogonal or orthographic ; in the former the projectors
"are inclined to plane of projection, in the latter they are
perpendicular. When the word "projection" is used with-
out qualification, orthographic projection is to be understood.
Thus the projection of a point on a plane is the foot of
the perpendicular let fall from the point to the plane.
Referring again to Figs. 1 6 1 (a), 1 6 1 (/>), the points a, a, a"
are seen to be the three projections of A on the planes of
reference. The projection a on the horizontal plane is called
the plan ; the two on the vertical planes are called elevations :
a is the front elevation, and a" the side elevation. In
Fig. 161 (a) the points is said to be represented by its
projections. The planes of reference are also called planes
of projection.
The lines aa and aa" in Fig. 1 6 1 (a) are also called pro-
jectors. They are perpendicular respectively to O Fand OX.
We may point out that in all cases the three projections show which
trihedral angle contains the point. Thus in Fig. 162 (,/) the point A
is behind the plane of YZ, making the x co-ordinate negative.
The point is projected on the planes of reference at a, a', a". Now in
order to derive Fig. 162 (b), the vertical planes of Fig. 162 (a) must be
rotated about the axes of A" and Y respectively, and always in the
same directions. These directions of rotation are those given when
describing the construction of the model, and are such as to open out
the positive dihedral angle. The angle in which the point lies and
its exact situation are given without ambiguity by Fig. 162 (b).
Notation. A capital letter is used to denote the point in space,
and the corresponding small letters denote the projections. Thus for
the point A the plan is a, the front elevation a , and side elevation a".
vii POSITION IN SPACE DEFINED AND EXHIBITED 1C3
Z,X'
a;
!/
n
\a'.
m.
162(a)
X 162(b)
ZX
Examples. 1. The x, y, 2 co-ordinates of a point A are
respectively 3", 4", and 2" ; draw the three projections of the
point and set up the model for this case.
2. Draw the projections of the three points (2.5", 3", 4"), ( 3",
4", 2"), and (3", - 4 ", -2").
3. In which trihedral angle is the point A situated whose projec-
tions are given in the figure below ? Measure the co-ordinates of
the point, scale th. Ans. -1.6", -3.6", -3".
Note. OX', OY', OZ',
OZ' represent the
negative directions of
the axes. 01, 0/n,
On are the x, y, z
co-ordinates of A,
and are seen to be all
set off in the negative
directions along their
respective axes, from
which we infer that
the three co-ordinates
are all negative.
Or thus Set up the model and in some way represent the
point A in its actual position, say by using a lady's hat-pin.
Look vertically downwards on the model ; the plane ZOY
appears as the line OY, and because the //an a is behind Y'OY
the point A must be behind the plane ZOY. Also the plane
ZOX appears as the line OX, and since the plan a is to the left
of OX, the point A is to the left of the plane ZOX. Now view
the model in the direction at right angles to the plane ZOY;
the plane XOY appears as the line OY \ and because the front
elevation d is below OY the point A is below the plane XOY.
Hence the point A as represented by its projections a, a', a"
is in the back-left- lower trihedral angle, and its co-ordinates are
found on measurement to be as given above.
I
,a"
y
Y
z
a'.
71
."
}ro
Z'
z;x
1 84 PRACTICAL SOLID GEOMETRY chap.
163. Problems on the position of a point. We suggest
here some simple problems, and give some examples for
the student to work out himself. It is intended that the
drawing-paper shall be cut, indented, and folded as
explained in Art. 161, and then fixed to the board with the
vertical planes YZ, ZX free to be laid flat while drawing,
or lifted into position while studying the problems, as often
as may be required. All the problems reduce to one or
other of the solutions of the right-angled triangle specified
in Art. i 2.
We give only hints. If formal solutions were written we
should defeat the object in view, which is to familiarise
the beginner with the method and signification of projections,
by encouraging the use of models.
From the lower figure it is seen that the point A and
its projections a, a', a", the points /, m, n, and O form the
eight corners of a rectangular prism, and OA is a diagonal
of the solid. Diagonals of some of the faces of the prism are
shown by dotted lines. We now suggest some problems,
having given as data the co-ordinates of the point A.
(a) To find the distances of A from the axes.
The distance of A from OX is equal to .41 or Oa ', that is to the
hypothenuse of the right-angled triangle Oma', the two sides of which
are two of the given co-ordinates.
(b) To find the distance of A from the origin.
Here we must find the length of the hypothenuse OA of the right-
angled triangle Oa'A, one side Ad being the given x co-ordinate, and
the other side Od being known from the above.
(c) To find the angles which OA makes with the -planes
of reference. (See Def. 8, Appendix II.)
The inclination of OA to the horizontal plane is the base angle
AOa of the right-angled triangle AOa. The inclination of OA to the
front, vertical plane YZ is the base angle AOa of the right-angled
triangle AOa'.
(d) To find the angles which OA makes with the axes.
The inclination of OA to the vertical axis OZ is the vertical angle
A On of the right-angled triangle A On. This angle is the complement
of the angle AOa.
(e) To draw the three projections of OA.
These are the lines Oa, Od, and Oa".
vii POSITION IN SPACE DEFINED AND EXHIBITED 1S5
7.
163(b)
163(a)
ft
a'
/
/
/
;
77b
z
ru
Y
I
X
c
v *
i
X
Examules. 1. The x, v, s co-ordinates of a point A are respec-
tively 3", 4", and 2".'
(a) Draw the three projections of A, viz. a, a', and a".
(b) Draw the three projections of OA, and measure their
lengths. A)is. 4.47", 3.O1", 5" on the planes of YZ,
ZX, X V respectively.
(C) Determine and measure the distances of A from the axes
of A', J', and /.. Ans. 4.47", 3.61", 5.0". .
186 PRACTICAL SOLID GEOMETRY chap.
(d) Determine and measure the distanceof A from the origin O.
A us. 5.39".
(e) Determine and measure the angles which OA makes
with the planes of projection. A /is. ^j.S , 48. o,
21.8 with the planes of YZ, ZX, X Y.
(f) Determine and measure the angles which OA makes with
the axes of A', J', and Z. Ans. 56. 2, 42.0, 68.2.
(g) Determine and measure the true distances of the projec-
tions a, a, a" from one another. A us. a" a = 4.47" ;
fl a' = 3.6i"; dd'=$".
(h) Set out to scale the true shapes of the four triangles
aa'a", Oa"a, Oaa', Odd'.
Note. In g and h the word true signifies that the results are to be
obtained for the true positions of the planes of projection, and
not for their positions when laid flat.
2. The three projections a, d, a" of a point A are given in Fig.
163 (a). Copy this figure double size, then measure the co-
ordinates, and obtain the results c to h of Ex. 1.
3. Draw the figure of Ex. 3, Art. 162, full size, then obtain the
results of Ex. 1 above.
164. Polar co-ordinates of a point. The position of a
point in space might be defined in other ways, e.g, by its dis-
tances from three fixed points. But the only two systems of
co-ordinates in general use are the rectangular and the polar.
In the polar method we choose a point called the
pole, an axis OZ through the pole, and a plane ZOX con-
taining the axis. The polar co-ordinates, say (r, 6, 4>) of
any point A are then (1) the polar distance OA or r ; (2)
the angle ZOA or 6 ; and (3) the angle between the planes
of ZOA and ZOX, say <.
In illustration, consider how a place on the earth's
surface is located. In the figure is the centre of the
earth, OZ the north polar axis, DXE the equator, ZOX the
meridian plane through Greenwich, intersecting the surface
in the meridian circle ZGX and the plane of the equator
in OX. Let A be the place on the surface, and let the
meridian plane ZOA intersect the surface in the meridian
circle ZAM, and the plane of the equator in OM. Then
the position of A is usually defined by its longitude, that is,
the angle XOM, and its latitude MO A.
vil POSITION IN SPACE DEFINED AND EXHIBITED 187
Choosing OZX for reference, the polar co-ordinates
(r, d, 4>) of the point A would be respectively the earth's
radius OA, the aTlatitude ZOA {i.e. 90 - latitude), and the
longitude XOM.
In the figure the polar co-ordinates of A are rOA,
6 = angle ZOA, and </> = angle XOM.
The problems resolve as before into cases of the
solution of the right-angled triangle.
Examples. 1. The rectangular co-ordinates of a point A are 3",
4", and 2" ; find the polar co-ordinates (r, d, 0).
Ans. 5.39", 6S.2 , 53. i.
2. The polar co-ordinates ;-, d, <p of a point A are respectively 3", 40,
and 55. Draw the projections of the point, and measure its
rectangular co-ordinates. Ans. 1.11", 1. 58", 2.3".
3. Find the polar co-ordinates of the point A, represented in projec-
tion in Fig. 161 (a), drawn | size. Ans. 5.46", 70.7 , 66. 5 .
4. Taking the latitude and longitude of Rome as 41. 9 N. and
1 2. 5 E., represent the position of Rome by a plan and two ele-
vations. Radius of the earth 4000 miles. Scale i"to 1000 miles.
5. A line 3-i" long from the origin to a point A makes angles of
40 and 6o respectively with the axes of X and Y. Draw the
projections of the line, and measure the rectangular and polar
co-ordinates of the point A. Ans. 2.6S", 1. 75", 1.4 1" ;
3.5", 66.2% 33. i.
iSS PRACTICAL SOLID GEOMETRY chap.
165. The straight line. In Fig. 165 (/>), let AB be a
straight line in space, and ab, a'b', a'b" its three projec-
tions. Then these projections, given as in Fig. 165 (a),
completely represent and define the line.
Or these projections could be drawn, if as data we were
given the co-ordinates of the two ends of the line.
Again arrange the drawing-paper so that the planes YZ, ZX
may be lifted into the vertical position at any time. The line AB
in space may be exhibited in the model by cutting a piece of paper
to the shape ABba and attaching it to the plane XY, as shown by a
folded margin left for the purpose ; in setting out this shape observe
that ab is equal in length to the plan of the line, and the perpendiculars
a A, bB are equal to the z co-ordinates.
Examples. 1. The rectangular co-ordinates of two points A and B
are respectively 2", l", {", and \\ 3", if".
(a) Draw the three projections of AB, and measure their
lengths. Ans. 2.24", 1.8", 2.5", on YZ, ZX, XY.
(b) Determine and measure the actual length of AB.
Ans. 2.69".
That is, find the length of the hypothenuse of a right-
angled triangle, of which the base is equal to the plan
ab, and the height to the difference of the 2 co-ordinates.
(C) Determine and measure the angles which AB makes
with the planes YZ, ZX, XY. Ans. 33. 9 , 48, 21. 8.
That is, find the angles between AB and its three pro-
jections a'b', a"b", ab. These all reduce to finding the
base angles of right-angled triangles.
(d) Determine and measure the angles which AB makes with
the axes of X, Y, Z. Ans. 56. i, 42, 68.2.
Since Aa is parallel to OZ, the angle which AB makes
with OZ is equal to the angle aAB. Similarly for the
other two angles. These angles are determined as the
vertical angles of right-angled triangles, and are com-
plementary to the angles of (c) above.
(e) Determine the projections of the points where AB pro-
duced meets the planes of projection YZ, ZX, X Y.
Measure the co-ordinates of the three traces so found.
Ans. (o, 3.67", 1.83"), (2.75", o, o), (2.75", o, o).
(f) Determine the true shape of the triangle OAB, and
measure the angle A OB. Ans. 72. 8.
First find the lengths of the sides, and then construct
the triangle.
vii POSITION IN SPACE DEFINED AND EXHIBITED iSg
z
n
at
lesta^
4
i
i
/
f
r
b'
\n
r
rrv
Y
Z ,j
OL L.
L
a,
9
>c "
y
-^^^ ^"-^
^"^^. ^^v
b
P
X
Copy Fig. 165 (a) double size. Then measure the co-ordinates
of A and B, and obtain the results b to /of Ex. 1.
Find the length of the line which joins two opposite corners of
a building brick 9" x 45" X 3". And determine the angles which
this line makes with the edges and faces of the solid. Ans.
Length =lo|". Angles with the 9", 4V', 3" edges = 31.0',
64.7 s , 73-4 Angles with the small, middle, and large faces
= 59.0, 25.3', 16.6 .
190 PRACTICAL SOLID GEOMETRY chap.
166. The plane. A plane, of indefinite extent, is located
when the positions of any three points in it, not in one
straight line, are known. In the rectangular system, the
points A, B, C, where the plane meets the axes, are the
three most convenient points to choose for the purpose
of defining the position of the plane. The lengths OA,
OB, OC are called the intercepts of the plane on the
axes. We may denote the lengths of these intercepts by
a, b, and c. The lines BC, CA, AB, where the plane
intersects the planes of projection, are called the traces of
the plane, and serve very well to represent the plane by
projection.
Thus, a plane is very cox\\zme\\\\y defined in the rectangular
system by its intercepts, a, b, c ; and represented in projectio?i
by its traces, BC, CA, AB
To the data of Ex. I below, let the student make the following
model. Cut out in paper a triangle whose sides are equal to AB, BC,
CA, Fig. 1 66 (a), leaving a folding margin along AB for attachment
to the horizontal plane as shown in Fig. 166 {/>) ; on it draw CD per-
pendicular to AB. Next draw OD perpendicular to AB and cut out
in paper a right-angled triangle, having OD for base, and OC for
height, leaving a margin along OD to be attached to the horizontal
plane, underneath the plane ABC, as shown in the lower figure. The
first example should now be worked by the student himself, without
assistance other than that given in the notes.
Examples. 1. The intercepts a, b, c of a plane on the axes of
X, Y, and Z are respectively 4", 5", and 3".
(a) Draw the three traces of the plane AB, BC, CA, and
measure their lengths.
Ans. 6.4", 5.83", 5"-
(b) Find the rabatments of the triangle ABC into the three
planes of projection.
A plane is said to be rabatted when it is turned about its
trace into the plane of projection. Thus ABC is rabatted
into the horizontal plane by rotation about AB ; and
into the planes of YZ, ZX, by being turned about BC
and CA respectively.
(C) Determine the rabatments of the triangle COD into each
of the three planes of projection.
VII POSITION IN SPACE DEFINED AND EXHIBITED ini
z
/
c
166(a)
/
/
/
/
/
1
I
c\
^\-#
z \
\ \
jr+ys
X
_/i/e/'
166(b)
y
(d) Determine and measure the angles which the plane ABC
makes with the planes of YZ, ZX, XY.
Ans. 5 7; 2, 64. 4 , 43. S.
The inclination of the plane ABC to the horizontal plane
is measured hy the angle ODC o{ the right-angled triangle
ODC. Corresponding constructions are required for
the inclinations to the other two planes of projection.
See Def. 9, Appendix II.
192 PRACTICAL SOLID GEOMETRY chap, vii
(e) Determine and measure the angles which the plane makes
with the axes of X, Land Z.
Ans. 32.8 , 25-6, 46. 2 .
These angles are complementary to the angles of (e).
(f) Draw the three projections of OP, the perpendicular,
from the origin to the plane. Determine and measure
the true length of this perpendicular.
Ans. 2.17".
The point P is in CD, and OP is perpendicular to CD.
(g) Find and measure the angles which the perpendicular
OP makes with the planes YZ, ZX, XV.
Ans. 32. 8, 25. 6, 46. 2.
(h) Find and measure the angles which OP makes with the
axes of X, Y, and Z.
Ans. 57.2 , 64.4 , 43.S .
Note the illustrations of the theorem, that the angle
between two planes is equal to the angle between any
two lines which are perpendicular to the planes.
2. Copy Fig. 166 {a) accurately, double size ; then measure the
intercepts and obtain the results b to g of Ex. j.
3. The latitude and longitude of a place A are 52 N. ; and
those of a place B are o and 40 W. (a) Find the angle sub-
tended by the two places at the centre of the earth. (/>) Find
the length of the shortest path on the earth's surface between
the two places. Radius of earth = 4000 miles, (c) Find the
direction of this path at each place. Ans. (a) 62. 4 . (b) 4410
miles, (i) At A, 46. 5 W. of S. ; at B 28 E. of N.
4. The latitude and longitude of Rome are respectively 41.9 N.
and 12.5 E., and of St. Petersburg 60. o N. and 30.3 E.,
obtain the results of Ex. 3. Ans. (a) 20 ; (b) 1440 miles.
(r) At Rome, 27.4 E. of N. ; at St. Petersburg, 43 W. of S.
CHAPTER VIII
FUNDAMENTAL RULES OF PROJECTION
167. Object of chapter. In this chapter we recall cer-
tain fundamental rules and methods of projection, developed
in Part I., and which the reader will have learnt from his
previous study. A collection of elementary examples is
then given to test the knowledge of the student. These
should present no difficulty. A well-trained student may
generally omit them and pass on at once to the next chapter.
168. The two principal planes of projection. Since the
form of a simple solid is often definitely shown by two pro-
jections only, the side vertical plane and side elevation of
the last chapter may generally be dispensed with. There
remain the two principal planes, the horizontal and the
vertical planes of projection. Their intersection is called
the ground line and will be denoted by XY ox xy. These
planes, extended both ways,
divide the neighbouring space -E E
into four dihedral angles.
A model should be made as
follows : Take two pieces of stout
drawing-paper, each about 9" by 6".
Cut one along AB and the other along
CD, CD. The latter is then folded along the lines DE, DE, passed
through the slit AB of the former, and unfolded. The planes may then
be turned into position at right angles to one another.
O
194
PRACTICAL SOLID GEOMETRY chap.
169. Situation in the four dihedral angles. Four points
situated respectively in the four angles are exhibited by
projection in the lower figures. These are derived as
indicated in the upper figures, by first projecting the plan
and elevation, and then turning the planes into coinci-
dence, about XY, so as always to open out the front upper
angle.
We may imagine either that the vertical plane has been
turned into the horizontal plane so that its upper portion
moves backwards, or that the horizontal plane has been
turned into the vertical plane, its front half falling. The
result is the same either way.
From the lower figures we can infer the positions of the
points. That is, we "can measure how much the points are
above or below the horizontal plane, and how much in front
of ox behind the vertical plane.
The following conception is useful in the interpretation
of drawings.
When considering an elevation, picture the drawing as
coinciding with the vertical plane of projection, and think
of the ground line xy as being an edge ox profile view of the
horizontal plane, just as if the model of the planes were
held on a level with the eye and viewed directly from the
front. We thus at once recall the rule
Rule 1. A point A is situated above or below the
horizontal plane according to whether its elevation a is above
or below xy. And the distance of A from the plane is
equal to the distance of the elevation a' from xy.
When considering a plan, picture the drawing as now
coinciding with the horizontal plane, and conceive the
ground line to be an edge view of the vertical plane, as if
the model of the planes were viewed directly downwards
from above. This at once leads to the rule
Rule 2. A point A is situated in front of or behind
the vertical plane according to whether it's plan a is in
front of or behind xy. And the distance of A from the
vertical plane is equal to the distance of the plan a from xy.
viii FUNDAMENTAL RULES OF PROJECTION 195
s
Y
i^\^
\
(4)
x
\a!
(L
y
X
b
W
y
X
\c
y
X
d
u-
y
This method of reading a drawing is often conducive to
clearness of conception, and should be cultivated by the
student. Instead of the planes of projection being imagined
as turned into coincidence, they are conceived as retaining
their horizontal and vertical positions, and it is the drawing-
paper which is supposed to be brought to coincide first
with one plane, then with the other, accompanied by a cor-
responding change in the direction of view.
Or thus, if we are reading the plan and elevation of any
solid object, and trying thus to imagine the form of the
latter, we think of the object as retaining its upright position,
and regard the elevation as a picture of the object when
viewed horizontally from the front, and the plan as its
apparent shape when viewed vertically downwards from
above. Thus we may regard xy as an elevation of the
horizontal plane, or a plan of the vertical plane, or as the
line of intersection of the two planes.
We add a third rule.
Rule 3. The projector drawn between the plan and eleva-
tion of a point is always perpendicular to xy.
196 PRACTICAL SOLID GEOMETRY chap.
170. Rules for drawing" auxiliary projections.
Definition i. An auxiliary elevation is the projection on
any vertical plane not parallel to the principal vertical plane.
This is illustrated in Fig. (i), where a" is the auxiliary
elevation of A on the auxiliary vertical plane X' Y' .
The lower figure exhibits this by projection, and the
perspective view indicates how the projection may be
derived by turning the two vertical planes backwards into
the common horizontal plane.
Observe that na" = ma' = aA, and that aa" is perpen-
dicular to xy . Hence the theorem :
Theorem i. If tivo or more elevations of a point be pro-
jected from one plan, the distances of the several elevations
from their respective ground lines are the same in all.
The following simple and effective
model should be made by all students.
Indent and fold a piece of drawing-
paper about 10" x 8" to represent the
principal planes of projection, and attach
an auxiliary plane, say 6" x 4", by paper-
fasteners through folded margins. See the
figure. The projections of A may be
drawn on this model.
Definition 2. An auxiliary plan is the projection on any
plane which is perpendicular to the vertical plane and not
parallel to the horizontal plane.
Fig. (2) illustrates this case. The perspective view shows
the point B projected on the three planes, and the arrow-
heads indicate how two planes may be turned into the
common vertical plane so as to obtain the lower figure.
Observe that nb 1 = ml? = b'B, and that b"b 1 is perpendicu-
lar to x x y v which leads to
Theorem 2. If two or more plans or a point be pro-
jected from the same elevation, the distances of the several
plans from their respective ground lines are the same in all.
The previous model will serve to illustrate this case, if it be held in
the proper position.
The projections having been drawn on it, the model may be viewed
VIII
FUNDAMENTAL RULES OF PROJECTION
197
at right angles to the auxiliary plane, or either hinge may be unfastened
for the purpose of rabatment.
Theorems 1 and 2 apply to any object, conceived as an
assemblage of points. They may be put in the form of
rules. Thus suppose the plan and elevation on xy to have
been drawn ; then
Rule 1. To obtain the auxiliary elevation on a neiv
ground litre x'y', project from the plan perpendicular to x'y,
and for the new elevation mark off on the projectors the dis-
tances of the points above {or beloiv) x'y equal to the distances
of the corresponding points of the old elevation above (or below)
xy.
Rule 2. To obtain the auxiliary plan on a ?iew ground
line x x y v project from the elevation perpendicular to x x y v and
for the new plan mark off on the projectors the distances of
the points in front of (or behind) x x y v equal to the distances of
the corresponding points of the old plan in front of (or behind)
xy.
198
PRACTICAL SOLID GEOMETRY
CHAP.
X
171. Sectional projections. An object is often assumed
to be cut in two by a section plane,
in order the better to show its
internal form. A projection of
either portion on the cutting
plane, or one parallel to it, is
called a sectional projection, the
severed parts being indicated by
section lines.
The figure shows the plan
and a sectional elevation of a
regular tetrahedron.
Note. The height of the solid is
found by rabatting the sloping edge AD
about its plan.
172. Developments of surfaces.
The surfaces of all polyhedra, and also certain curved
surfaces, including the cone and cylinder,
can be developed; that is, unfolded into
one plane without any wrinkling or
stretching.
A workman who makes objects of
sheet metal, or a student who makes
paper models of geometrical solids, first
cuts out the shape of the development.
The illustration shows a develop-
ment of a regular octahedron, with a margin left for
up the edges of a paper model of the solid.
173. Figured projections. The
diagram shows the projection of an
irregular tetrahedron, in which the
numbers affixed to the letters at the
corners indicate to a given unit or
scale the distances of the points from
the plane of projection. In this
way one projection is sufficient to
define a solid form.
gluing
Unit =0-1"
viii FUNDAMENTAL RULES OF PROJECTION 199
174. Miscellaneous Examples.
1. Explain in your own words the meaning of the terms projector,
projection, plan, elevation, section, planes of projection.
2. Represent in plan and elevation four points A, B, C, and D,
situated respectively in the four dihedral angles, each point
being iV distant from the horizontal plane, and 2^" from the
vertical plane.
3. Draw the plan and elevation of a point A which is in the hori-
zontal plane, and 1" behind the vertical plane. Also of a point
B, in the vertical plane, and 2" below the horizontal plane.
4. A point 1.72" below xy is both plan and elevation of a points ;
in which of the four dihedral angles is A situated ? Find the
true distance of A from the ground line. Ans. 2.43".
5. Draw the plan and elevation of a point 1" below the horizontal
plane, and 2" distant from xy. How many solutions are there ?
Ans. Two.
6. Determine the plan and elevation of a cube of 2^" edge :
(a) when one face rests on the horizontal plane, and an edge
makes an angle of 25 with xy ; (b) when an edge rests on the
horizontal plane perpendicular to xy, and a face is inclined at
an angle of 65 to the horizontal" plane.
7. A cube of 2" edge is pierced by holes 1" square through all its
faces, so as to form a framed or skeleton cube. Draw the
plan and elevation when in the positions of Ex. 6.
8. Draw the plan and elevation of a square pyramid, side of base
2", length of axis 2 -J" : (a) when the base rests on the ground
with one side making an angle 30 with xy ; (b) when a
triangular face rests on the ground, the axis of the solid being
parallel to the vertical plane.
9. A point A is I-J-" above the ground, and 2" in front of the vertical
plane. Determine the auxiliary elevation of A on a vertical
plane which makes an angle of 50 with the vertical plane of
projection.
10. ABC is a triangle, the heights of A, B, and C above the
ground being 1", 2", if", while the corresponding distances
from the vertical plane are 1", 2J", f". In plan ab measures
2" and be if". Draw an elevation of the triangle on a
vertical plane which makes 50 with xy.
11. A point A is 1" above the ground and 2" in front of the
vertical plane. Draw the auxiliary plan of A on a plane at
right angles to the vertical plane of projection and inclined
at 6o to the horizontal plane.
12. ABCD is a square. The three corners A, B, C are 2", lV,
and l" above the ground, their distances in front of the
vertical plane being ih", 2", I J". In plan ab is 2" and be
200 PRACTICAL SOLID GEOMETRY ciur.
i" long. Draw the plan and elevation of the square, and an
auxiliary plan on a plane at right angles to the vertical plane,
and parallel to AD.
13. A regular tetrahedron rests with one edge (2" long) on the
ground, whilst a face containing that edge is inclined at 40 .
Show in plan the section of it made by a horizontal plane 1"
high.
14. A prism, 3" long, the ends of which are equilateral triangles
of 1^" side, rests with a rectangular face on the ground. Draw
the plan and a sectional elevation on a vertical plane which
bisects the axis of the prism at an angle of 45. Draw a
development of the portion of the surface of the solid situated
on one side of the cutting plane.
15. A cube, 2" edge, rests with one edge on the ground and at right
angles to the vertical plane of projection. A face containing
this edge is inclined at 25 ; draw the plan and elevation of
the cube. Draw also an auxiliary elevation on a vertical plane
which is parallel to a diagonal of the solid.
16. Draw a sectional elevation on a vertical plane which bisects that
edge of the cube in Ex. 15 which is on the ground, the cutting
plane making 70 with the principal vertical plane.
17. A hollow sphere, 2^" external diameter and i|" internal
diameter, has a portion cut away by a plane distant J" from the
centre. Draw a sectional projection of the larger remaining por-
tion, on a plane which divides it into two exactly equal parts.
18. A cube 2" edge is pierced centrally by holes |" square through
all its faces. It rests with one face on the ground. Draw a
sectional elevation on a vertical plane which contains a vertical
edge and bisects a horizontal edge of the cube.
19. A hexagonal pyramid, edge of base 1", height 3", rests with the
base on the ground, one side of the base making io with xy.
Draw the plan and elevation, and an auxiliary plan on a plane
perpendicular to the vertical plane and parallel to a long edge
of the pyramid.
20. Suppose the pyramid in Ex. 19 to rest with a side of the base
on the ground and perpendicular to xy, the base being inclined
at 25 to the ground. Draw the plan and elevation, and a
sectional plan on a plane which passes through the centre of
the base, and is parallel to a long edge of the pyramid.
21. Draw a line ab 2" long, and attach indices of 18 and 25 to a
and b respectively, thus obtaining the figured plan of a line
AB. Find the true length of AB by drawing an elevation
on a vertical plane taken parallel to AB. Unit = o. 1".
If ABC be an equilateral triangle having its plane vertical,
determine the indexed plan of C,
vin FUNDAMENTAL RULES OF PROJECTION 201
22. Copy the figure of Art. 173 three times the size, keeping the
indices the same. Draw an elevation of the pyramid on a
vertical plane parallel to AC. Draw also a sectional elevation
of the pyramid on a vertical plane through A and the middle
point of CD. Index the plan of this section.
23. Draw a quadrilateral, abed, making ab 2J", be 2", the angle
abc 120 , cd 4", ad 3^"; choose v within abed, and such
that av is if", and vd 2^". Attach indices of 12, 8, 18, 24,
and 35 to a, b, e, d, and v. Join v to a, b, e, d; the result is
the figured plan of a pyramid with vertex r 'and base ABCD.
Determine (a) the plan of a section by a horizontal plane, at a
level 19 ; (/') a sectional elevation on a vertical plane, bisecting
AVxdA CV. Unit 0.1".
24. Draw the plan of a cube, 2" edge, when a diagonal of the
solid is (a) vertical, (/') horizontal.
25. A building brick has a line joining two opposite corners
of the solid vertical. Draw a sectional plan on a hori-
zontal plane passing through the centre of the brick. Size
of brick 9" x 4" x 3". Scale .
26. An instrument box with the lid open at an angle of 120 rests
on the ground ; draw its plan and a sectional elevation on a
plane parallel to one end. Draw also an elevation on a vertical
plane which makes 45 with the plane of one end.
Dimensions of box outside length 6", breadth 4V',
depth of box 1^", depth of lid f". Thickness of wood at
sides and ends |-", at top and bottom j". Scale ^.
27. Draw a development and make a paper model of each of the
following six solids (a) a cube, (b) a regular tetrahedron, and
(e) a regular octahedron, each of 2" edge ; (d) a building brick
9" x 4V x 3", scale J ; (e) a prism 3" long, ends equilateral
triangles, 2" side ; (/) a pyramid 3" high, base a square of 2"
side.
28. Draw plan and elevation of a regular tetrahedron, 2" edge,
when resting with one edge on the ground and at right angles
to the vertical plane, a face containing that edge being
vertical.
29. A regular octahedron, i^" edge, rests with a face on the
ground. Draw its plan and elevation, when one edge of that
face is parallel to the vertical plane. Draw a sectional eleva-
tion on a vertical plane which bisects any two adjacent hori-
zontal edges.
30- A building brick 9" x 4^" x 3" is cut into two unequal portions
by a plane which contains three corners. Draw the plans of
the two parts when resting on their section faces. Index the
plans of the corners of the solid.
202 PRACTICAL SOLID GEOMETRY chap.
*31. A solid is formed of two equal square prisms side of base
lh", height 3V the axes of which bisect each other at right
angles. The elevation is shown but is not drawn to scale.
Draw this elevation the proper size, and deduce the plan, and
a second elevation on a new ground line, making 50 with
xy. (1S78)
*32. The figure is the elevation of a truncated hexagonal pyramid.
Draw the plan of this solid inverted, so that the plane of
truncation is on the horizontal plane of projection. (1889)
* 33. The position and dimensions of two equal and equally inclined
rectangular prisms are indicated by their given projections.
Draw the sectional elevation on AB. (1 877)
*34. The figure represents an end elevation of an open trunk. The
length is twice the breadth. Draw a front elevation on a
plane parallel to one of the diagonals of the bottom. The
thickness of wood may be neglected. (1884)
*35. A spherical segment rests on the top of a truncated hexagonal
pyramid. The plan and elevation are partially given. Draw
the figure four times the given size, and make a sectional
elevation of the solid on AB. (1888)
"36- The plan and elevation of a desk are given. Draw a front
elevation on a line parallel to x^y v (1886)
*37. Make a sectional elevation on AB of the desk. (1886)
38. Draw the plan of a square prism, height 2", side of base
\\", a diagonal being vertical. ( 1891 )
39. ABC is the base of a pyramid, and Fits vertex. AB = 2,",
AC=z", BC=2, AF=CV=3", BV=zV. Draw th
e
plan of the pyramid when A is ih", B2h", and C 1" above the
horizontal plane. (1896)
viii FUNDAMENTAL RULES OF PROJECTION 203
CHAPTER IX
THE STRAIGHT LINE AND THE PERPENDICULAR PLANE
175. The possible positions of a line. The various
positions which a line, situated in the first dihedral
angle, may have, relatively to the planes of projection, are
shown lettered (a) to (n).
The remaining lines (o) to (/) are wholly or partially in
one or more of the other three angular spaces.
The lines (a) and (b) are respectively perpendicular -to
the horizontal and vertical planes; (c) is parallel to the
vertical plane and inclined to the horizontal plane ; (d)
is parallel to horizontal plane and inclined to the vertical
plane ; and (e) is parallel to both planes.
The line (/) is in the vertical, plane ; (g) is in the
horizontal plane ; and {h) is in both planes.
The lines (k), (/), (m), (n) are inclined to both planes,
(/) having one end in xy, (m) being perpendicular to xy,
and (n) having its two ends respectively in the planes of
projection.
In studying the position of the lines, let the student
take the model described in Art. 168.
Pieces of wire passed through the paper, or held in the
hand, may be used to represent the lines, and the wires
should be placed in positions so as to agree with the pro-
jections given.
When the model is held with the horizontal plane on
CHAP. IX THE STRAIGHT LINE AND PLANK
205
(a)(b) CO (d) (e) (f) (g) Qv) (k)
\y/b <4>
bb'
CC
< F
'& a b
ou
aa.
(l)(m){rv) (0) (p)
(?) (T) (s) (t)
a level with the eye, and viewed from the front, the lines
should appear respectively as the elevations a'b' ; when
looked down upon vertically, the appearance should be
that of the plans ab.
Examples. 1. Show in plan and elevation : (a) a point 2" from
the ground line and ij" from the vertical plane of projection ;
(b) a line parallel to and ij" from the vertical plane of pro-
jection, and inclined at 40 to the horizontal plane.
2. Draw the projections of any four lines, which are situated wholly
in the four dihedral angles, one in each.
3. Show by their projections any two equal parallel lines not
parallel to either plane of projection.
4. How can you tell from the projections of two lines whether the
lines meet or not ? Draw the projections of any two lines which
intersect, and of two lines which do not intersect.
5. A line 2" long is parallel to xy, and distant 2" therefrom ; is
this sufficient data to fix the position of the line ? Draw the
projections of the line, if it be 1" in front of the vertical plane.
206 PRACTICAL SOLID GEOMETRY chap.
176. Problem. Having given the plan ab and eleva-
tion ab' of a line not parallel to either plane of projection,
to determine the length of the line, and its inclinations,
a and /?, to the horizontal and vertical planes respect-
ively.
There are three useful methods of solving this problem.
First Method. By auxiliary projections, Fig. (i). Take
x'y parallel to ab, and draw the auxiliary elevation a"b".
Also take x l y 1 parallel to ab', and draw the auxiliary plan
Then both a x b x and a"b" give the length of AB. The
angle marked a is the inclination of AB to the ground ; and
the angle marked /3 the inclination to the vertical plane.
Second Method. By rabatments, Fig. (2). Draw aA ,
bB perpendicular to ab, and equal to ma, nb' respec-
tively. Draw a A v b ' B x perpendicular to ab' , and equal
to ma, nb respectively. Then A B is the rabatment of
AB about ab into the horizontal plane ; and A X B X is the
rabatment of AB about a'b' into the vertical plane.
The length of the line AB is either A Q B Q or A X B X ; and
the inclinations are the angles marked a and f3.
A model should be made, by cutting out the shapes of the two
quadrilaterals in paper, leaving a margin along the edges ab and a'b' ,
for attachment to the planes of projection.
Third Method. By turnings about projectors, Fig. (3).-
With centre b, radius ba, describe the arc aa { , to meet
ba x drawn parallel to xy. Project from , to meet the hori-
zontal through a' in /. Then a'b' AB, and b'a l 'a' = a.
With centre a', radius a'b', describe the arc b'b } ', to meet
the horizontal through a' in b x . Project from b( to meet
the line through b parallel to xy, in b y Then ab^ = AB,
and ab 1 b = fi.
The first construction represents AB turned about Bb, so that AB
becomes parallel to the vertical plane. In the second construction AB
is turned about Aa , until AB is horizontal.
The lower figure shows three additional examples of this
problem, all solved by the method of rabatments.
IX
THE STRAIGHT LINE AND PLANE
207
Examples. 1. The plan and elevation of a line are 2" and 3"
long. The projectors of its extremities are l" apart. Find
the true length and inclinations of the line. Ans. 3.46",
54-7, 6o.
2. One end of a line is .75" below the horizontal plane ; the other end
is 1.5" above it. What is the true length and inclination of the
line if its plan measures 2" ? Ans. 3", 48 .
3. The elevation of a line 3" long measures 1.5" ; at what angle is
the line inclined to the vertical plane ? Ans. 6o.
2oS PRACTICAL SOLID GEOMETRY chap.
177. Problem. Having given the projections of a line
AB, to find the horizontal and vertical traces H and V.
The plans and elevations of the traces must be respect-
ively in the plan and elevation of the line (produced if
necessary) ; also the plan of the vertical trace and the
elevation of the horizontal trace must both be in xy. The
construction is therefore as follows :
Produce the elevation to meet xy in h' ; from h' draw a
projector to meet the plan produced in //. H is the hori-
zontal trace required.
Produce the plan to meet xy in v, draw vv to meet the
elevation a'b' produced in v'. V is the vertical trace of the
line.
In the right-hand figure the horizontal trace is behind the
vertical plane. The perspective diagram underneath illus-
trates this case.
The construction given in this problem fails when the
line is perpendicular to xy. In this case, find the rabat-
ments of the line, and, if necessary, produce these to meet
the projections produced, as shown in Fig. 6, page 207.
When a line is parallel to a plane, its trace on that plane
may be considered at an infinite distance away in the
direction of the line.
178. Problem. Having given the indexed plan of a
triangle, to find the true shape. Unit = 0.05".
Let abe, with the indices, be the given plan.
Determine the rabatments of the three sides, as in the
second method of Prob. 1 76, and draw the triangle A B Q C
with its sides respectively equal to these rabatments.
A B C is the true shape of the triangle, and may be
regarded as being the rabatment of the triangle into the
horizontal plane first, by turning the quadrilateral BCcb into
the ground about be, then by further turning the triangle
into the ground about -E> C .
Examples. 1. The plan of a line is 2" long and its elevation 3".
The projectors of its extremities are 1" apart. If the lower
IX
THE STRAIGHT LINE AND PLANE
209
end of the line be 1" in front of the vertical plane and 1^"
above the ground, determine the horizontal and vertical traces
of the line.
Regarding the lower end as fixed, how many lines are there
satisfying the above data ? Ans. Four.
2. Draw a triangle abc having ab=2\", be =2", ac=7,". Attach
indices of 6, 15, 24 to a, b, and c respectively. Determine
the true shape of the triangle ABC by finding the true lengths
of its sides. Unit = 0.1".
Ans.- A B = 2.66", BC=2.jg", CA = $.$".
3. Determine d U) in ac of Ex. 2. Join d and b, then DB is a hori-
zontal line at a level 1 5. Now draw x'y at right angles to db, and
obtain an auxiliary elevation of the triangle on x'y' ; this should
be a straight line, the edge elevation of the plane of the triangle.
This is a useful construction.
210 PRACTICAL SOLID GEOMETRY chap.
179. Solutions depending on the right-angled triangle.
We have seen that many problems in practical geometry
reduce to problems on the right-angled triangle. The nine
cases of the latter have already been given in Art. 12, but
the importance of the recognition of these analogies seems
to warrant a repetition of the cases specially applicable to
the problems under immediate consideration.
Let OPp, OQq be two right-angled triangles. Then
if OP be the true length of a line, and a its inclination to
the ground, Op is the length of its plan, and Pp the
difference in the vertical heights of its ends. Any two of
these four quantities being given, the other two may be
found by constructing the triangle OPp.
Similarly, if OQ be the length of a line, (3 the inclination
to the vertical plane, Oq is the length of its elevation, and
Qq the difference in the distances of the ends of the line
from the vertical plane. Any two of these quantities being
given, the remaining two can be determined.
180. Problem. A line 2" long has one end V' high ;
the plan of the line measures 1\", and the elevation If".
Draw the projections of the line.
Determine by the method of Art. 179 the difference
in the heights of the ends of the line, and in the distances
of the ends from the vertical plane. That is, in Fig. 179
find Pp and Qq, having given OP= OQ 2", op=\h",
oq = if". Then proceed as follows :
Draw any line ab, 1 |" long, as the plan. With centre b,
radius Qq, describe a circle. From a draw am, a tangent
to the circle, and take xy parallel to am. Then the
elevation on xy is the one required.
In drawing the elevation, a is to be set off |" above
xy, and b' a distance Pp (Fig. 179) higher than a. The
length of the elevation will then be found to be if",
because the difference in the distances of A and B from
the vertical plane, being equal to bm, is equal to Qq
(Fig. 179).
IX
THE STRAIGHT LINE AND PLANE
211
S^< ^7 \
180 v n
/
./
Examples. 1. Take a point a in xy, and draw ab I j" long, perpen-
dicular to xy. Let ab be the plan of a line AB, of which B
is I V higher than A ; find the length and inclination of AB.
If A be Jf " above the ground, draw the elevation of AB, and
find the horizontal trace of the line.
2. A point A is 2" above the ground and 1^" in front of the vertical
plane. AB is a line 2|" long, whose plan is perpendicular to
xy ; draw the plan and elevation of AB, when B is in the
vertical plane.
3. The plan of a line measures 2" and the elevation 1^". Is this
data sufficient to fix the length of the line? If not, add any
third condition which will fix the length.
A is a point in the vertical plane ij" above the horizontal
plane ; B is a point in the horizontal plane i|" from the vertical
plane. The real distance from A to B is 3". Draw the plan
and elevation of the line AB.
4. A point is 2" from the vertical plane and 1.75" above the hori-
zontal plane. Determine a point B on the ground line 3j"
from A, and measure the inclination of AB.
5. Two points A and B are respectively 1.5" and 1.75" from
both planes of projection. Their plans are 2.25" apart. De-
termine the projections of a point C in the vertical plane 2" from
A and 2.75" from B. If C has to be 2" from A, what is the
least possible distance of C from B? Ans. 2".
6. Determine the projections of a point D in AB, Ex. 5, distant
1.75" from A.
7. A line AB is inclined at 35". The upper end A is 2" above the
ground, and ab is 2" long. Draw the elevation of AB on x'y'
which makes 30 with ab.
8. A line 3" long has one end 1" high, the plan of the line measures
2" and the elevation 2j". Draw the projections of the line.
212 PRACTICAL SOLID GEOMETRY chap.
181. Problem. A line AB of given length has its ends
in the planes of projection, and the line is inclined at
given angles a and 6 to the horizontal and vertical planes
respectively. Draw the plan and elevation.
Method i. By the construction of Art. 179 this problem
could be at once converted into the case of the preceding
problem, and thus solved.
Method 2. Draw B x d and B x c inclined at a and /3 to xy,
and make B x d = B x c = AB. Draw the projectors da, cc.
Then, as in Art. 179, B x a is the length of the plan of
the line, and B x c the length of the elevation.
Make db' = B x c' , and from b' draw a projector to meet
the circle, with a as centre, and aB x as radius, in b. Then
ab, db' are the required projections.
To explain this construction, suppose in the first instance the line to
be in the vertical plane, so that aB v a B x are its projections. Let the
line be then turned about ad as axis. The path of B is a circular arc
of which B x l> is the plan, and B x b' the elevation. The rotation is
continued until the length of the elevation is equal to B x c ; the line is
then inclined at /3 to the vertical plane. It is also inclined a to the
horizontal plane, since the latter inclination remains unchanged during
the rotation.
Limits to data. The sum a+/5 cannot be greater than 90 .
182. Problem. Having given the projections of a
line AB, to find the shortest distance of the line (produced
if necessary) from xy.
Let ab, db' be the given projections.
Draw 0J1 and o v, an edge view of the horizontal and ver-
tical planes of projection.
Determine a Q , the projection of A in this view, by
making a m = a'/, that is, equal to the height of A above
the ground; and a Q n = a/, that is, equal to the distance of
A in front of the vertical plane.
Similarly determine b and join a Q > . Draw o^ Q perpen-
dicular to a b . Then o c is the shortest distance required.
Obtain c from c , and c from c ; then co, c'd are the
projections of the shortest line between xy and AB. This
line is perpendicular to both xy and AB.
IX
THE STRAIGHT LINE AND PLANE
213
Examples. 1. Draw the plan and elevation of any line 2\" long,
which is inclined at 35 to the horizontal plane, and 25 to the
vertical plane.
2. A line 2^" long has one end in xy, and is inclined at 30 and
40 respectively to the horizontal and vertical planes of projec-
tion. Represent the line.
3. Draw the plan and elevation of any line 3" long, which shall be
equally inclined to the planes of projection, and not parallel
to xy.
4. The plan of a line 2.\" long measures 2" ; draw the elevation on
a vertical plane inclined to the line at 20 .
5. The plan of a line measures 1" and the elevation .8". The in-
clination of the line to the horizontal plane is 30 ; what is the
inclination to the vertical plane? Ans. 46. 2.
6. Draw a triangle abc having ab = 2^", ac = \\
7.
8.
9.
be
Through
draw cd 3" long, cutting ab and making an angle of 50
with ca. Attach indices of 8, 18, 6, 15 to a, b, c, d.
Find whether the lines AB and CD intersect. Unit = o. 1".
If they do not, find what the index of d must be so that
the lines shall intersect. Ans. 25.6.
Draw the plan of a horizontal line at a level 10 to intersect
the lines AB, CD in Ex. 6.
A point A is Y above the ground, and 1^" behind the
vertical plane. Find the distance of A from xy. Ans. 1. 58".
A point A is 1" above the ground, and ii" in front of the
vertical plane. A point B is h" above the ground, and 2|" in
front of the vertical plane. The projectors of the points are
1 f" apart. Find the shortest distance between AB and xy.
Draw the projections of the shortest line between AB and xy.
214 PRACTICAL SOLID GEOMETRY chap.
183. The perpendicular plane. A perpendicular plane is
one which is at right angles to one or other (or both) of the
principal planes of projection ; a plane which is inclined to
both being called oblique. A perpendicular plane may be
either inclined or vertical. An inclined plane is one which
is inclined to the horizontal plane and perpendicular to the
vertical plane. A vertical plane is perpendicular to the
horizontal plane and may be inclined to the vertical plane
of projection.
As was stated in Art. 166, a plane is conveniently repre-
sented by its traces. We shall now have only two traces,
the horizontal and the vertical.
The student should make a model of a perpendicular plane as shown
at (a). For the planes of projection take a piece of drawing-
paper 9" square, indented and folded across the middle. For the per-
pendicular plane, cut a rectangular piece of drawing-paper 6|" by 5j",
and indent and fold margins |" wide on two adjacent edges for attach-
ment by paper-fasteners. By taking out one or other pair of fasteners,
the plane can be rabatted into either plane of projection at pleasure ;
and by placing the model the two ways up in succession, it illustrates
both the inclined and vertical plane, as shown at (a) and (/').
The characteristic feature of a perpendicular plane is
that one or other of its traces is a profile or edge view of the
plane. In the oblique plane neither trace is so.
To illustrate this let the student take the model, and
holding it in the position (a), let him view it directly from
the front ; the whole plane will appear as a line coinciding
with the vertical trace. The vertical trace is thus strictly
an elevation of the plane, for it contains the elevations of all
points and lines in the plane. The horizontal trace could
not be called a plan of the plane, for it contains the pro-
jection of only one line in it, the trace itself, as will be quite
evident from the model.
Similarly in (/>), it will be at once seen from the model
that the horizontal trace is a true plan of the vertical plane,
for it contains the plans of all points and lines in the plane.
But the vertical trace is not an elevation.
IX
THE STRAIGHT LINE AND PLANE
215
v v
V
-v
h
(a) (7? j (C)
(d)
y
h
Cej
Five perpendicular planes are represented by their traces
in the lower figure : (a) as an inclined plane, (b) a vertical
plane, and (c) a plane perpendicular to xy ; (d) is a hori-
zontal plane, and (e) a plane parallel to the vertical plane
of projection. In (a) tv is an edge elevation of the plane ;
in {b) th is a profile plan ; and in (1), (</), and (e) the traces
are all edge views of the planes.
By using the model the student will see that the angles
which the planes make with the horizontal and vertical
planes of projection are respectively equal to the angles
which the vertical and horizontal traces make with xy. In
an oblique plane this is not the case.
It will also be noticed that the apparent angle between
the traces is not the real angle, for the latter is a right
angle.
216 PRACTICAL SOLID GEOMETRY chap.
181 Problem. A square pyramid is given by its
projections, (a) Determine the plan of the section made
by any inclined plane VTH. (b) Draw the plan of the
frustum with its section end on the ground, (c) Draw a
development of the surface of the pyramid, showing the
trace of the cutting plane on the surface.
(a) The elevations of P, Q, P, S, the points in which
the plane cuts the edges, are on tv, because the vertical
trace is an edge elevation of the plane.
The plans/, q, r, s are obtained by projection from the
elevations.
(b) On tv draw an auxiliary plan of the frustum, as
shown in the figure ; this is determined according to the
second rule of Art. 170, and is the plan when standing on
the section end.
The plan may also be considered as obtained by rabat-
ment of the plane about its vertical trace, the frustum
having been first projected on the section plane.
(c) To obtain a development, first find the true length
of a sloping edge of the pyramid and the true distances of
P, Q, P, S from the vertex. In the figure v'd^ and v'q(
are the true lengths of VD and VQ respectively, obtained as
in Prob. 176, third method. The lengths of FP, VP,
VS may be found in a similar manner.
A development of the surface of the pyramid must now
be drawn as shown, and the lengths VQ, VP, VS, and
VP set off along the developed edges. The development
is then completed by joining QP, PS, SP, PQ, these lines
forming the developed trace.
Examples. 1. A cube, 2" edge, has its base on the horizontal
plane, one corner being in xy, and the diagonal of the base
through that corner making 6o with xy. Draw the plan and
elevation showing the section by a plane passing through the
centre of the cube, and inclined at 30 . Find the true shape of
the section, and draw a development of the frustum.
2. A regular tetrahedron ABCD of 3" edge rests with ABC on the
ground. Draw its plan and show the section by a plane through
the centre of the solid, whose horizontal trace is am making
45 with ab.
IX
THE STRAIGHT LINE AND l'LANE
217
-y
b
218 PRACTICAL SOLID GEOMETRY chap.
185. Problem. Having given the projections of a cube,
and the traces vth of a vertical plane, to determine the
plan, elevation, and true shape of the section.
The horizontal trace lit is an edge view of the plane, and
the required plan of the section is therefore in the horizontal
trace, and is the line pq. The elevation of the section is
the shaded rectangle obtained by projection from pq.
The true shape of the section is the rectangle pQ m
obtained by the rabatment of the section plane into the
horizontal plane, pP being made equal to p'p'.
The true shape might be equally well obtained by a
rabatment into the vertical plane, about the vertical trace,
and the student should make this construction.
Example. A square pyramid, side of base i|", axis 2^", has its
base in the vertical plane, one corner being in xy, and the diagonal
of the base through that corner inclined at 6o. Draw the plan
and elevation, showing the section by a vertical plane, bisect-
ing the axis of the solid, and making 30 with the vertical
plane of projection. Find the true shape of the section.
186. Problem. Having given the plan of any polygon
or plane figure lying in a given inclined plane, to find the
elevation, rabatment, and true shape of the figure.
Let abc be the given plan, and vth the traces of the
given plane.
From the points in plan draw projectors to meet the
vertical trace in a, b', c . Then db'c is the required
elevation of the polygon. For, the vertical trace being an
edge elevation of the plane, it must contain the elevation of
the polygon.
The true shape may be obtained by a rabatment of the
plane about its horizontal trace. With centre / describe the
arcs a'A ', b'B ', c'C '; these are the elevations of the
paths of A, B, C; and A ', B \ C ' are the elevations of
the rabatments of A, B, C. The plans of the arcs are the
lines aA , bB , cC , drawn perpendicular to the horizontal
axis of rotation th. These intersect the projectors from
A ', ', Q in A v B , C .
IX
THE STRAIGHT LINE AND PLANE
219
>>.v
% \C y
The true shape may also be obtained by a rabatment
into the vertical plane ; this is equivalent to an auxiliary
plan on tv, and may be determined by applying the
principles of Art. 170.
This is an important problem because it occurs so
often in other problems ; hence the student should make
himself quite familiar with the construction.
Examples. 1, Draw an equilateral triangle abc, 2" side, with ab
making an angle of 45 with xy. Let this be the plan of a
triangle ABC lying in a plane inclined at 40 . Find the in-
clinations of the sides of the triangle.
2. By using the rabatment of the triangle ABC in Ex. 1, determine
(a) the plan of the bisector of the angle ACB ; (/') the plan
of the perpendicular drawn from C to AB.
Note.
-Observe that if in the rabatment the angle A C Q B n be
bisected by a line which meets the horizontal trace of the plane
in ?', then ic is the plan of the bisector. In problems in-
volving rabatments it is a very useful artifice to produce lines
to meet the trace in stationary points.
220 PRACTICAL SOLID GEOMETRY chap.
187. Problem. To find the point of intersection of a
given line AB and inclined plane VTH.
The elevation of the required point of intersection is
/', for reasons stated in previous problems. The plan p is
found by drawing the projector from /' to meet ab.
188. Problem. To determine a perpendicular from a
given point A to a given inclined plane VTH.
Draw a'm perpendicular to tv ; through ni draw the
projector mm, to meet at m a line through a perpendicular
to th. Then AM is the required perpendicular.
The student should use the model to illustrate this problem.
189. Problem. To determine the plan and elevation of
the projection of a given point A on a given inclined
plane VTH.
Proceed as in the last problem. Then M is the required
projection.
190. Problem. Having given a line AB and an inclined
plane VTH, to determine (a) the trace of the line on the
plane ; (b) the projection of the line on the plane ; and
(c) the angle between the line and plane.
(a) The required trace of the line on the plane is the
intersection S, and is found as in Prob. 187.
(b) The projection Jl/JV 'of the line on the plane is de-
termined by applying the construction of Prob. 189.
(c) The inclination of the line to the plane is the angle
between the line AB and its projection MN on the plane,
and is found by a double rabatment as follows :
Fust, MqNq, the rabatment of MN, is found as in Prob.
186. Then A B Q JV Af Q , the rabatment of the quadrilateral
ABMJV about M N Q , is determined by drawing M A , N B Q
perpendicular to M Q 2V Q and equal to am, b'ri, the true
lengths of the perpendiculars AM, BN.
The inclination of the line to the plane is the angle
AqSqMq.
The student may readily make a model to illustrate this problem,
by cutting out the shape of AqBqMqN'q in paper, with a margin at
MqNq for attachment to the model of the inclined plane.
IX
THE STRAIGHT LINE AND PLANE
221
TTL
188
Oj
k
222 PRACTICAL SOLID GEOMETRY chap.
191. Problem. Determine the plan and elevation of a
line inclined at a, and lying in a plane inclined at e.
Find the angles which the line makes with the traces of
the plane, and the inclination of the line to the vertical
plane.
Draw the traces vth of the plane inclined at 9.
Take any point A in the vertical trace of the plane ; its
elevation a will be in tv, and its plan a in xy.
Through A draw a line in the vertical plane inclined at
a, with one end B on the ground, so that a'B 1 is its eleva-
tion, and a 1 its plan.
Let this line revolve about ad until its end B comes into
the horizontal trace of the plane ; the line itself is then in
the plane, because / both its ends are in it. The arc B Y b
struck with a as centre is the plan of the path of B, and
B x b' is the elevation of the path.
Then ab, db' are the projections of the line in the re-
quired position ; for the inclination of the line remains un-
changed during the rotation.
Let bA be the . rabatment of the line obtained as in
Prob. 1 86, then the true angles which AB makes with the
horizontal and vertical traces are respectively A bb' and
bA b\ The latter angle is also /i, the inclination of AB to
the vertical plane, since it is equal to the angle between
the line and its elevation.
This problem is an important one, and should be studied
with the model, until the exact meanings of the five angles
referred to are fully understood, and the differences between
them are recognised. Students partially acquainted with
the construction are apt to take wrong centres for the
two arcs.
Limits to the data. The angle a cannot be greater than 6, but may
have any value from O to 6. When a = o, the line is horizontal, and
therefore parallel to the horizontal trace of the plane. When a = 0,
that is, when the line and plane are equally inclined, the plan of the
line is perpendicular to the horizontal trace, and the line itself in the
plane is also perpendicular to the horizontal trace.
IX
THE STRAIGHT LINE AND PLANE
22'
./
191
192
192. Problem. Determine the plan and elevation of a
line inclined at 6 to the vertical plane, and lying in a
plane inclined at e to the horizontal plane. Find a, the
inclination of the line to the ground.
Let vth be the traces of the plane inclined at 0.
First draw bA Q making an angle j3 with xy, and suppose
this to be a line in the horizontal plane.
Now let this line be turned about bt until it comes into
the plane VTH. That is, with centre / describe the arc
A a\ and draw the projector a a.
Then ab, a'b' are the projections of the line in the
required position.
The inclination a is obtained by turning AB into the
vertical plane, about a a, as in Prob. 176, third method.
The construction of this problem corresponds to that of
the preceding problem worked backwards, and the two
problems should be studied together with the model.
Example. A vertical plane makes 45 with xy. Draw its traces
and show a line lying in it and inclined at 45. Find the
angle which this line makes with the vertical plane of projection.
224 PRACTICAL SOLID GEOMETRY chap.
193. Plane represented by a scale of slope. It lias
already been explained that by a system of indexed plans,
points and lines may be represented by one projection
only. It has now to be shown how in the same system a
plane is represented.
Draw a line through any two points, P, Q, on a piece of cardboard,
and let one edge A B of the cardboard be perpendicular to PQ. Take
the cardboard as the model of a plane, and hold it in an inclined
position, with AB resting on any flat surface representing the horizontal
plane.
The student will then observe that PQ is inclined at
the same angle as the plane, and that the plan of PQ is
perpendicular to the horizontal trace AB. Also that any
line parallel to PQ, and lying in the plane, has the same
inclination as the plane.
Lines in the plane such as PQ, perpendicular to the
horizontal trace, are called lines of slope, and it is evident
that if the figured plans of P and Q, or of any two points
on any line of slope, be given, the position of the plane
relatively to the horizontal plane is completely defined.
The figured plan of a line of slope is called a scale
of slope, and is conventionally drawn as two lines, one
thicker than the other, as is usual in drawing scales. The
double line serves to distinguish the representation of a
plane from that of a line.
Fig. 193 shows the representation of a plane by a scale
of slope, the unit for the indices being 0.1".
To represent this plane in the manner previously
explained, see the next problem.
194. Problem. Having given a line by its indexed
plan, a 12 b 5 . 5 , and a plane by its scale of slope, 5,15, to deter-
mine the indexed plan of the point of intersection of the
line and plane.
Draw any xy parallel to the scale of slope, and find
db', the elevation of AB ; find also 5', 15', the eleva-
tions of the two given points on the scale of slope.
IX
THE STRAIGHT LINE AND PLANE
225
193
Then vt, drawn through 5', 15', is the vertical trace of
the given plane as regards xy ; and p' is the elevation of
the intersection of the line and plane, as in Prob. 187.
Project from /' to determine the plan p. To find
the index of p, measure the height of/' above xy, or read
the position of q on the scale of slope.
Example. Draw a line ab 1^" long ; draw ad and be each perpen-
dicular to ab and I" and i"in length. Attach indices of 10, 30,
25, and 15 to a, b, c, J. Regard ab as the scale of slope of a
plane and cd the figured plan of a line. Determine the indexed
plan of/', the intersection of the line and plane. Unit 0.1".
195. Problem. Having given the indexed plan of a
triangular pyramid, and the scale of slope of a plane, to
determine the indexed plan of the section of the solid.
(No figure.)
Draw an elevation of the plane and pyramid on an xy
taken parallel to the scale of slope, and proceed as in
Prob. 184. Measure the heights of the points in the
section, and index the corresponding points in plan.
Q
226 PRACTICAL SOLID GEOMETRY chap.
196. Miscellaneous Examples.
1. Draw the traces of a perpendicular plane, inclined to the right at
40 to the ground ; draw ab 1" long anywhere to the right of
the horizontal trace. Regard ab as the plan of a line AB lying
in the plane, and determine the plan of a square ABCD also
lying in the plane.
Hint. First find the rahatment A B C D ; produce C B to
meet the horizontal trace in i ; join ib and produce it to meet in c
a line from C parallel to xy ; then complete the parallelogram
abed. Observe that by working with i we do not require to
use the elevation in order to determine c from 6? .
2. Represent two planes inclined, one at 40 and the other at 70 ,
and such that their intersection is I J" from the ground. Obtain
the projections of a horizontal line, i" above the ground and
2V' long ; the ends of which are in the two planes. Determine
the point of intersection of this line with the inclined plane
which bisects the acute angle between the first pair of planes.
3. Determine the traces of a plane inclined to the right at 15 ,
and perpendicular to the vertical plane. A point A is \\" in
front of the vertical plane, and 2" above the horizontal trace,
but 1" to the right of it. Draw the projections of the per-
pendicular let fall from A on to the plane.
4. Represent a vertical plane making an angle of 40 with xy.
In this plane place a line 2j" long, which shall be inclined to
the vertical plane of projection at an angle of 30 .
5. Represent a vertical plane which makes an angle of 40 with.rj', and
draw the projections of a square lying in this plane, one diagonal
being inclined at 70 , the lower end of which is on the ground.
6. An inclined plane makes an angle of 40 with the ground ;
draw the projections of a line which bisects the angle between
the traces.
7. Draw the plans of two lines AB, AC, which lie in a plane
inclined at 50 , the lines being inclined at 30 and 45
respectively. Obtain the projections of the bisector of the
angle BAC.
8. Draw the traces of a plane inclined at 50 , and in this plane
place a line 2" long inclined at 40 . Find the inclination
of the line to the vertical plane.
9. The horizontal trace of a vertical plane makes 42 with the
ground line. Determine the elevation of a line lying in this
plane, inclined at 30, and passing through the point where the
given plane cuts the ground line.
10. Draw the traces of a vertical plane which makes an angle of 6o
with the vertical plane of projection. Represent a line lying
in this plane, and inclined at 70 to the vertical trace.
ix THE STRAIGHT LINE AND PLANE 227
11. A person on the top of a tower 60 feet high, which rises from
a horizontal plane, observes the angles of depression of two
objects A and B on the plane to be 20" and 30, the directions
of A and B from the tower being west and south respectively.
Find (a) the distances of A and B from the tower ; (/>) the
distance apart of A and B ; and (c) the direction of B and A.
12. Represent a plane inclined at 50 , and in it place two lines,
one horizontal and the other inclined at 30 . Find the true
angle between these lines.
13. The face of a hill is inclined at 30 , the lines of slope being
due east ; what is the inclination of a path on the hillside which
goes in a north-easterly direction.
14. A plane inclined at 40 contains a point A, distant 1" and ih"
respectively from the horizontal and vertical planes of projec-
tion. Determine the rabatments of the point about the traces
of the plane.
15. A line is inclined at 35 and 50 to the horizontal and vertical
planes respectively, its traces are 4" apart. Determine its
projections.
16. A line ab 2" long, making 30 with xy, is the plan of a
horizontal line 1 j" above the ground. Find its elevation, and
determine the plan and elevation of an isosceles triangle, having
the given line for base, and its vertex in xy.
17. Two points are respectively 1.5" and 0.75" from both planes
of projection. Their plans are 2.25" apart. Determine a
point on the ground line equidistant from the points.
18. A rectangle, sides 2" and 3", revolves upon one diagonal as a
fixed horizontal line until the plan of a right angle opposite
becomes 120 . What is then the inclination of the plane of
the figure ?
19. Two lines inclined to the horizontal plane at angles of 25 and
45 respectively are drawn from a point situated 2 " from both
planes of projection. The plans of these lines make 110
with each other. Determine the real angle between the
lines.
20. Given a point 1.25" from the horizontal plane, and 1" from the
vertical plane ; obtain the projections of any line 4-I" long
passing through the given point and terminated by the planes
of projection.
21. A line is inclined at 30 and 40 to the planes of projection ;
what is its inclination to a plane which is perpendicular to xy ?
22. Draw the plan of a regular octahedron of 1 h" edge when resting
with one face on the ground. Determine the plan and true
shape of the section made by a plane which is inclined at
45 and contains one diagonal of the solid.
228 PRACTICAL SOLID GEOMETRY chap.
""23. Determine the inclination of the line AB (Fig. a) to each plane
of projection.
K "24. Determine the horizontal traces, //"and K, of any pair of lines
which pass through P (Fig. a), and intersect AB in, say, C and
D. Show in the same figure the true shapes of the triangle
PHK and the quadrilateral CD// A'.
*25. From a given point/,/' (Fig. a), draw a perpendicular on the
given line ab, a'b'. ( 1 877)
Hint. Proceed as in Ex. 24, and having obtained the true
shapes PqH^A'q and C D H A' , draw a line from P perpen-
dicular to CqDq, meeting A/ A" in Z . The projections of the
required perpendicular may be at once obtained from Z Q .
*26. A line parallel to the vertical plane is given by its projections
ab, a'b' (Fig. b). Draw the traces of a plane containing this line
and perpendicular to the vertical plane. In this plane draw
lines passing through A and B and making 45 with the line
AB. Draw also an elevation of the three lines on a vertical
plane perpendicular to the ground line. (1885)
*27. a'b' (Fig. c) is the elevation of a line 2!" long. Its centre C
is 2" from xy. Determine (1) the plan of C, (2) the difference
of the distances of A and B from the vertical plane of
projection.
*28. a'b' (Fig. c) is the elevation of a line 2i" long. Its centre
point is 2" from xy. Draw its plan. (1884)
*29. (Fig- d) is the lowest corner of a cube ; ab is the plan of
one edge, AB (real length = 20 units), of a face which lies in
a plane, of which ht is the horizontal trace. Draw the plan
of the cube. Also the plan of its section by a plane parallel
to the horizontal plane of projection, and at a height of 12
units above it. Unit = o. 1". (1896)
*30. Determine the height of P, a point on CD, whose figured plan
is given (Fig. <?). Unit = 0.05".
*31. In Fig. (e) p is the centre of a sphere 2\" radius. Determine
a sectional elevation of this sphere on a vertical plane whose
horizontal trace is ab. Unit 1=0.05".
""32. Two lines are given by their figured plans, Fig. (e). From
the point P on one of them draw a line 2 -J" long, terminating
on the other. Unit = 0.05". (1886)
Hint. Proceed as in Ex. 31, obtaining an elevation of AB
on the vertical plane whose horizontal trace is ab.
*33. Find the length of the line CD in Fig. (/), and determine the
length of the diagonal of a cube of which CD is one edge.
Unit = 0.1".
*34. Determine whether the triangle CDE, Fig. (/), is right-angled
at D.
IX
THE STRAIGHT LINE AND PLANE
229
CHAPTER X
THE OBLIQUE PLANE
197. Introduction. As defined in Art. 1S3, an oblique
plane is one which is inclined to both planes of projection.
The traces are both inclined to xy, or both parallel to it, or
both coincide with it ; and the true angle between the
traces is not a right angle.
The figure shows four cases of an oblique plane. In
(a) the real angle between the traces is acute ; in (b) it is
obtuse ; in (c) the traces are parallel to one another ; and in
(d) the plane contains the ground line, and cannot be repre-
sented by its two traces in the ordinary way, since these
both coincide with xy. It requires to be shown by a side
elevation, on the plane perpendicular to xy, as indicated
in the figure.
Let the student illustrate these cases by four models, made in the
manner explained in Art. 183, and shown in the figure. The planes
are attached to the planes of projection by paper-fasteners passing
through folded margins ; either margin can be unfastened to allow of
the plane being rabatted about the other trace.
The student should note that the intersection of the
horizontal and vertical traces of a plane is always on the
ground line xy. When one trace is parallel to xy, so is the
other, and the intersection of the three lines is at an infinite
distance away in the direction of the ground line.
CHAP. X
THE OBLIQUE PLANE
2 3i
(a) &> h (b)
(C)
A characteristic feature which distinguishes an oblique
plane from one which is perpendicular to one or both of the
planes of projection, is that neither trace is an edge or pro-
file view of the plane. On this account the problems are
generally more difficult than those of Chap. IX. We shall
begin with some cases which may be made to depend on
the constructions of Chap. IX.
232 PRACTICAL SOLID GEOMETRY chap.
198. Problem. To convert a given oblique plane into
an inclined plane, by means of an auxiliary elevation.
In the diagram VTH \s, the given oblique plane.
Take Mn, a new vertical plane of projection, at right
angles to the given plane. Then X'Y,, the new ground
line, is perpendicular to TH, the horizontal trace.
With reference to this new plane of projection the given
plane is an inclined plane (Art. 183).
Let ON be the intersection of Mn and J'TH Then
ON is the new vertical trace, and is also an edge elevation
of the plane VTH.
The above process has now to be represented by ortho-
graphic projection.
Let vth be the given traces of the plane.
Take a new ground line x'y perpendicular to th, inter-
secting th in and xy in n. Draw nn, nn" perpendicular
to xy, x'y, and make nn nn. Thus n" is the auxiliary
elevation on x'y' of the point N. And on" is the new
vertical trace required.
The student should make a special model to illustrate this problem,
by cutting a model of the oblique plane along a line NO perpendi-
cular to TH, and inserting the new plane of projection Mn.
On holding the model so as to look along HT, it is
seen that X' Y is the new elevation of the ground, and NO,
the new vertical trace, is the edge elevation of the plane.
The lines nn, nn" are the positions taken by nN, the
line of intersection of the two vertical planes of projection,
when these planes are turned back about their respective
ground lines into the horizontal plane.
It should be observed that the angle v'on is the true
inclination, 6, of the plane to the ground.
Figs. (l>) and (c) show two other examples of this
problem. In Fig. (l>) the plane vth is the same as in Fig.
(a), but x'y' is drawn in a different position, to avoid the
overlapping of the two elevations. Fig. (r) shows what
form the construction takes when the real angle between
the traces is obtuse.
X
THE OBLIQUE PLANE
233
By means of the construction just given, many problems
on the oblique plane may be at once converted into cor-
responding problems on the inclined plane, and thereby
simplified. This method of attacking such examples is a
most valuable one, on account of its wide applicability,
and should become quite familiar to the student. The
six problems which immediately follow are worked in this
manner.
Example. Convert the following oblique planes into inclined
planes by means of auxiliary elevations, and in each case
measure the inclination to the horizontal plane :
(a) The horizontal and vertical traces ///, tv make angles of
40 and 55 with xy.
(/>) The traces make 120 and 35 with xy.
(<) The traces are in one straight line making 45 with xy.
Am. (a) 65.8 ; (6) 39^ : (<r) 54.7".
234 PRACTICAL SOLID GEOMETRY CHAP.
199. Problem. To determine the point of intersection
of a given straight line AB, and a given plane VTH.
Take x'y' perpendicular to th, and on x'y draw elevations
of the plane and line as shown.
The problem is thus reduced to that of Prob. 187,
c" being the auxiliary elevation of the point of intersection
of the line and plane.
The plan c is obtained by projecting from c" : and the
elevation c by projecting from r.
200. Problem. To determine the plan and elevation
of the section of a given prism by a given oblique plane
VTH. Also to draw a sectional plan of the solid.
By drawing an auxiliary elevation, this problem is con-
verted into that of finding the section of a solid by an
inclined plane, and is then worked as In Prob. 1S4.
Take x'y' perpendicular to tA, and on x'y' draw the
auxiliary elevations of the plane and prism as shown.
Then ov' is an edge elevation of the plane, and a ", />", c"
are the auxiliary elevations of the points where the plane
cuts the three vertical edges of the prism.
The heights of a', />', c above xy are now made equal to
the heights of </', />", c' above x'y', thus determining the
elevation of the section. The plan of the section is abc,
coinciding with the plan of the prism.
The required sectional plan is a projection of one portion
of the solid on the cutting plane, or on a plane parallel to it.
Take ov as a second auxiliary ground line, and from
the auxiliary elevation project the plan a^/ v making the
distances of a v /> v c x from ov the same as the distances of
a, b, c from x'y, according to the rules of Art. 170.
The sectional plan is completed by drawing the plans
of the other lines of the frustum, obtained in a similar
manner.
Compare this plan with the plan of the frustum of the
pyramid in Prob. 184. The two are obtained by the appli-
cation of the same principle.
X
THE OBLIQUE PLANE
235
/A
Examples. 1. Draw the traces of the three planes of the example
on page 233. Draw the projections of a line parallel to xy
and ii" from each of the planes of projection. Determine the
plan and elevation of the intersection of this line with each of
the three planes.
2. Draw the plan and elevation of a square pyramid, side of
base 2", height 3", the base being on the ground with its
centre if from xy, and one edge making 35 with xy.
Select a point on xy, 3-J" to the left of the plan of the centre
of the base. Through this point draw horizontal and vertical
traces making 45 and 38 respectively with xy. Determine
the plan and elevation of the section of the pyramid made by
this plane. Draw also a sectional plan of the pyramid.
236 PRACTICAL SOLID GEOMETRY chap.
201. Problem. To determine the perpendicular from
a given point A to a given oblique plane VTH.
Convert the problem into that of Prob. 188 by
drawing ov' and a", the auxiliary elevations of the plane
and point on x'y', taken perpendicular to th.
Through a" draw a" m" perpendicular to ov ; then a" m"
is the auxiliary elevation of the perpendicular.
Through m" draw the projector m"m to intersect in m
the line through a perpendicular to th.
The required plan, am, is thus determined ; and the
elevation on xy is at once found by projection from the
plan, since the height of M is known.
It may be proved by pure solid geometry that
Theorem. If a line and plane be perpendicular to each
other, the plan and elevation of the line are respectively per-
pendicular to the horizontal and vertical traces of the plane.
The student may satisfy himself of the truth of this
proposition by using a model. The accuracy of the above
solution should be tested by ascertaining whether am is
perpendicular to tv, as ought to be the case.
202. Problem. To determine a plane which shall bisect
a given straight line AB at right angles.
Take x'y parallel to the given plan ab, and draw the
auxiliary elevation d'b". Draw p"o bisecting a"b" at right
angles, and meeting x'y in o. Then p"o is an edge eleva-
tion on x'y, of the required plane.
The horizontal trace of the plane is ot, drawn through
o perpendicular to ab ; and the vertical trace is tv, drawn
through / perpendicular to ab'.
203. Problem. To determine the true distance be-
tween two given parallel oblique planes. (No figure.)
If the planes are parallel, their horizontal and vertical
traces are also respectively parallel (Theorem 13, Appendix).
The required distance between the planes is that be-
tween their edge elevations, obtained as in Prob. 198.
THE OBLIQUE PLANE
237
Examples. 1. Represent an oblique plane, and a line perpendi-
cular to it. Show also an oblique plane and a line lying in it.
2. Draw the traces of the three planes of the example on page 233,
and in each case draw the projections of a point A, situated 1"
to the right of /, 2j" above the ground, and 2" in front of the
vertical plane. Draw the projections of the perpendicular
from A on to each plane.
3. The horizontal and vertical traces of a plane make angles of 30
and 50 respectively with xy. Take a point in xy. 2" from
/, and determine the plan and elevation of the line OM. drawn
at right angles to meet the plane. Determine the angle which
OM makes with xy. Ans. 62. 3 .
4. A line AB 3" long has its ends I J" and 2^" from each of
the planes of projection. Determine a plane which shall
bisect AB at right angles.
5. Draw the traces of any oblique plane. Take any two points a
and b in the horizontal trace, and any point c' in the vertical
trace. Find the true shape of the triangle ABC.
6. Draw the traces of any oblique plane and those of
parallel to it, such that the perpendicular distance
the horizontal traces is 1". Determine the true
between the planes.
7. The horizontal traces of two parallel planes are \\" apart and
make 45 with xy ; the vertical traces are ft" apart, and the
angle between the traces k obtuse. Find the distance between
the planes. Ans. 6".
a plane
between
distance
238 PRACTICAL SOLID GEOMETRY chap.
204. Problem. Having given an oblique plane VTH,
and the plan of a point A in the plane, to determine the
projections of a straight line through A which shall lie in
the plane, and make a given angle d with the horizontal
trace.
Determine ov, the edge elevation of the plane.
Through a draw the projector to meet ov in a". Then
a" is the auxiliary elevation of the point A.
Obtain A , the rabatment of A about oh, as in Prob.
1 86. Through A draw^ ^, making the given angle 8 with
the horizontal trace oh. Then A Q b is the rabatment of the
required line.
When the plane goes back into its original position, the
point B does not move, being in the axis of rotation. The
plan of the required line is therefore ab, and the elevation
a'b' can be drawn since the heights of A and B are known.
205. Problem. Having given an oblique plane VTH,
and a point A in it, to determine the projections of a
straight line AB, which shall lie in the plane, and be in-
clined at a given angle a to the ground.
The line is first drawn through A parallel to the vertical
plane ; inclined at a ; its lower end on the ground. In this
position its projections are a'B\ and aB v
The line is then turned about Aa until B comes into the
horizontal trace of the plane. The line is then altogether in
the plane; it is still inclined at a; and it passes through^.
The required projections are ab, a'b' , the manner of
obtaining which is obvious. Compare Prob. 191.
206. Problem. To determine a straight line which
shall pass through a given point P, have a given inclina-
tion a, and be parallel to a given oblique plane VTH.
First, by Prob. 205, draw any line AB in the plane,
having the given inclination a. Then draw a line through
P parallel to AB.
The plan of the line is drawn through p, parallel to ab,
and the elevation through p' parallel to a'b' .
X
THE OBLIQUE PLANE
239
Examples. 1. The traces vt, th of a plane make angles of 45
and 30 with xy. Select a point a 2" to the right of / and 1"
below xy : this is the plan of a point A in the plane. Draw
the plan and elevation of a line AB 2^" long, which lies in
the plane and makes 30 with the horizontal trace.
Determine also the projections of a line AC 3" long, lying
in the plane, the angle BAC being 50'.
Hint. Having obtained A C , produce it to meet the
horizontal trace in /'. To find the plan c, join ia and produce
it to meet in c a line from C perpendicular to ///. If the point
i obtained as above be not within the limits of the paper,
draw
a more convenient point i, and
then c will lie on id produced.
2. The traces of a plane each make 50 with xy. A point A
in the plane is 1" from the vertical plane, and 2" above the
ground. Determine the projections of a line AB lying in
the plane, and inclined at 35 (a) to the horizontal plane, (b)
to the vertical plane. What is the greatest inclination which
AB may have either to the horizontal or vertical plane ?
3. Determine a plane which passes through the point A, Ex. 2,
and is parallel to the given plane.
e other line from C fl to meet the horizontal trace in
A B in Z) ,
from D obtain d,
2 4 o PRACTICAL SOLID GEOMETRY chap.
207. Problem. Having given the plan of any polygon
or plane figure lying in a given oblique plane, to find the
elevation and true shape of the figure.
Draw an edge elevation of the plane as in Prob. 198,
then proceed as in Prob. 186.
208. Problem. Having given a straight line and an
oblique plane, to determine (a) the trace of the line on the
plane ; (b) the projection of the line on the plane ; and
(c) the angle between the line and plane.
Draw an edge elevation of the plane, and a new eleva-
tion of the point on the same ground line. The problem
is thus reduced to Prob. 190, and is worked in the
manner there described.
209. Examples on Problems 198 to 208.
*1. Draw an edge elevation of the plane, Fig. (a), on a ground line
taken through b. Draw also an edge plan on x-,y-., taken
through a'.
*2. Fig. (/>) shows the plan of a prism with equilateral ends, resting
on the ground. Draw the elevation, and show the section by
the given plane. Draw also a sectional projection of the prism.
*3. Find the intersection of the line and plane in Fig. (e). Also in
Fig. (/)
_x "4. In Fig. (a) determine a perpendicular from A to the plane VTH.
*5. Determine a plane bisecting the line AB, Fig. (e), at right
angles.
*6. Find the distance between the parallel planes of Fig. (d). Show
the projections of any line perpendicular to the planes with one
end in each plane.
*7. In Fig. (a) the point B lies in the plane ; find its elevation.
Show a line containing B, lying in the plane, and making 70"
with the trace ///.
*8. Determine a line through B, Fig. (a), lying in the plane and
inclined at 30' 3 to the ground.
*9. In Fig. (a) determine a line containing B, lying in the plane and
making 50 with vt. Also one Inclined at 50 to the vertical
plane.
*10. In Fig. (<) the triangle whose plan abc is given lies in the
given plane. Determine the elevation and true shape of ABC.
*11. In Fig. (e) determine the trace of the given line on the given
plane. Show the projection of the line on the plane. Find
the angle between the line and plane.
*12. Obtain the results of Ex. 1 1 with reference to Fig. (f).
THE OBLIQUE PLANE
241
R
242 PRACTICAL SOLID GEOMETRY chap.
210. Problem. Having given an oblique plane VTH,
and one projection of a point A which lies in the plane,
to determine the other projection.
The method of solution is to first find the projections of
any line drawn through A and lying in the plane. Then
to draw a projector from the given plan to intersect the
elevation of the line in a.
( i ) Let the plan a be given.
In Fig. (a), through a draw be, the plan of any line
which has its lower and upper ends, B and C, respect-
ively in the horizontal and vertical traces of the plane ;
the elevation of the line is evidently b'c, B and C
being respectively in the horizontal and vertical planes of
projection. The projector from a, intersecting b'c in a,
gives a the required elevation.
In Fig. (b) the line through A is horizontal, so that
its plan ac is parallel to ///, and its elevation a'c parallel
to xy.
In Fig. (c) the line is taken parallel to the vertical
plane ; therefore its plan ab is parallel to xy, and its eleva-
tion a'b' parallel to tv.
(2) Let the elevation a be given.
The elevation of the line through A is first drawn ; the
plan of the line is then determined ; and finally the
required plan a is found by projecting from a '.
211. Problem. Having given the plan and elevation
of a point A, and one trace of a plane which contains A,
to find the other trace.
Let a, a, and th be given, Fig. (a).
Draw be through a ; project b' ; through c draw the pro-
jector to intersect b'a' produced in c . Then tc is the re-
quired vertical trace.
If the given trace be nearly parallel to xy, so that the
point / is not readily accessible, two lines such as BC
must be drawn through A, thus determining two points in
the other trace. See Fig. 215.
X
THE OBLIQUE PLANE
243
y (a) {b) h
210 and 211
Examples. 1. Represent a plane by its traces ; select a point a any-
where except in xy. Draw the plan of any line whatever which
lies in the plane and passes through A. Find the elevation of
this line, and by means of it find the elevation of A. Find the
inclination of the line through A which you have represented.
2. Draw a line making 40 with xy, and also the projections
of any point whatever. Determine the vertical trace of the
plane which contains the point, and has the first line for its
horizontal trace.
3. Draw a line inclined at 30 to xy, and draw the projections of
any point not in either of the planes of projection. Determine
the horizontal trace of the plane which contains the point and
has the first line for its vertical trace.
4. Draw the projections of any point. Take a point I ^". below xy,
and through it draw a line inclined at 5 to xy. Let this be
the horizontal trace of a plane containing the first point.
Determine the vertical trace.
Hint. Draw the plans of any two lines through the given
point. Regard these as lying in the plane ; draw their elevations
and their vertical traces, then join the latter. See Fig. 215.
We advise the student to remember this useful construction.
5. A point A is 1" above the ground and 1.8" behind the vertical
plane. A second point B is 1.5" below the ground and 0.5"
in front of the vertical plane. Determine a plane which con-
tains A and B and is parallel to the ground line.
6 3 Represent a plane which contains the ground line and also the
point A of Ex. 5. Find the angle between this plane and the
plane of Ex. 5.
244 PRACTICAL SOLID GEOMETRY chap.
212. Problem. To determine a plane which shall pass
through a given point A and be parallel to a given plane
VTH.
The student may easily work this problem by drawing
an edge elevation. The following is the special method :
Through a draw ac parallel to ///. This is the plan of
a horizontal line in the required plane, and a'c, the eleva-
tion of the line, is drawn as shown.
A point C in the vertical trace of the required plane
is thus found. Through c draw Im parallel to tv, and
through m draw mn parallel to th. The required plane is
LMN.
213. Problem. To find the rabatments of a given
point A which lies in a given oblique plane VTH.
First, to find A , the rabatment of A into the horizontal
plane, about th.
Draw am perpendicular to th. Then am is the plan of a
line in the plane, at right angles to the horizontal trace. It
is also, when produced, the plan of the circular path traced
by A during rabatment ; so that A must lie on this line.
Obtain a" am, the rabatment of the right-angled triangle
MAa about am ; observe that aa" = la' . Make mA = ma",
the true length of MA. Then A is the required rabatment
of A into the horizontal plane.
It will be noticed that the arc a"A is the rabatment
about aA of the path traced by A. As before stated,
aA is the plan of the path.
Next, to find A v the rabatment of A into the vertical
plane, about tv. The construction is similar to that just
described ; an is the elevation, and a x ri the rabatment
into the vertical plane, of a line AN lying in the plane at
right angles to the vertical trace ; n'A 1 is the rabatment of
the same line about tv ; and A 1 is the required rabatment
of the point A into the vertical plane.
A model to illustrate this important problem may be
made in the following manner :
THE OBLIQUE PLANE
245
Take a model of the oblique plane hinged along ///, and on it draw
the line AM perpendicular to th. Cut out in paper a right-angled
triangle of the shape a"ma, with a margin along am for attachment
to the horizontal plane. The two rabatments about am and th re-
spectively may then be effected.
This model may be used to illustrate the rabatment of a point into
the vertical plane, by turning it so that the horizontal plane becomes
the vertical plane, and vice versa.
Examples. 1. Draw the traces of any plane VTH, and also the
projections of a point A lh" vertically over ht and 2" from the
vertical plane. Determine a plane through A parallel to VTH.
2. The traces vt, th of a plane make 50 and 6o with xy.
Determine the rabatment, into the horizontal plane, of a point
A in vt and 2" from /.
3. Taking the plane of Ex. 2, determine the rabatment, into the
vertical plane, of a point A in ht 2" from /.
4. Determine the projections of a point A in the plane of Ex. 2,
2" from the vertical plane and 1^" from the horizontal plane.
Determine the two rabatments of A.
5. A plane at right angles to xy contains a point A which is 1"
above the ground and 2" from xy. Represent the plane and
point and determine the two rabatments of the latter.
6. A plane contains xy and the point -4 of Ex. 5 ; set out the two
rabatments of A.
246 PRACTICAL SOLID GEOMETRY chap.
214. Problem. To determine co, the true angle
between the traces of a given oblique plane VTH.
Take a, a, the projections of any point A in the vertical
trace of the given plane.
Obtain A , the rabatment of A about th, by drawing
aA perpendicular to ///, and making tA = ta .
Then tA is the rabatment of the vertical trace into the
ground, and the true angle between the traces is that
marked w. Two cases, (a) and (b), are shown.
215. Problem. Having given two intersecting
straight lines, to determine the plane containing them ;
also, to find the true angle between the lines.
In order that the lines shall intersect, the apparent
intersections in plan and elevation must lie on the same
projector.
The horizontal and vertical traces of the plane must
pass through the corresponding traces of the line.
Let the given lines be as shown in Fig. 215.
As in Prob. 177, determine H, A', the horizontal
traces, and V, S, the vertical traces of the given lines.
Then, lines drawn through //, k and v', s' are respectively
the horizontal and vertical traces of the required plane.
To find the true angle between the lines, obtain A , the
rabatment of A, as in Prob. 213. Join A /i, A k.
Then hAJz is the rabatment of the triangle HAK, and the
angle hA k is the true angle between the lines.
216. Problem. Having given two parallel straight
lines, to determine the plane containing them ; also, to
find the true distance between the lines.
Let the given parallel lines be as shown in Fig. 216.
Their plans and elevations are respectively parallel.
Determine the traces of the lines, through which draw
the traces of the required plane, as in the last problem.
The true distance between the lines is that between
their rabatments, which are shown dotted in the figure, and
are obtained as in the last problem.
THE OBLIQUE PLANE
247
216 \ *T
V *
Examples. 1. Determine the true angle between the traces of
each of the planes of the example on page 233. In each case
draw the projections of the bisector of the angle between the
traces. Also, draw the projections of a line equally inclined
to the traces and such that the part of the line intercepted by
the planes of projection is 3" in length.
2. Draw the projections of any two lines AB, AC. Find (a)
the traces of the plane containing the lines, (b) the inclina-
tion of this plane, [c) the angle BAC, (d) the projections
of the bisector of the angle BAC, (e) the projections of the
perpendicular from A on to BC.
3. Draw the projections of any two parallel lines. Determine the
plane containing them, and the distance apart of the lines.
248 PRACTICAL SOLID GEOMETRY chap.
217. Problem. To determine a plane which shall
contain three given points, A, B, C ; also, to find the true
shape of the triangle ABC.
Draw lines through any two pairs of the given points,
and then find the plane containing these lines, as in
Prob. 215.
To find the true shape of the triangle ABC, obtain the
rabatments of A, B, and C by the method of Prob. 213, or
work by an auxiliary elevation, as in Prob. 207.
218. Problem. To determine a plane which shall
contain a given straight line and a given point ; also, to
find the true distance of the point from the line.
Take any point in the given line and join it to the given
point. Then the plane through the two intersecting lines,
found as in Prob. 215, is the one required.
To find the true distance between the point and the
line, rabat the plane containing them, and draw the
perpendicular from the rabatment of the point to the
rabatment of the line.
219. Problem. To determine a line which shall lie
in the plane of two given intersecting straight lines, and
shall bisect the angle between them.
Draw the rabatment of the lines as in Prob. 215.
Then through A (Fig. 215) draw a line bisecting the
angle kA k, and intersecting hk in p (not shown).
Thus Ave obtain A p, the rabatment of the required
bisector, and ap its plan ; and since P is on the ground
the elevation of AP is known.
220. Problem.- To determine the angle between two
non-intersecting straight lines.
Through any point in one of the lines draw a line parallel
to the other.
The angle between these two intersecting lines, found
by Prob. 215, is the angle required (Def. 10, Appendix).
THE OBLIQUE PLANE 249
221. Problem. To find the angle between two given
planes.
Take any point A, and through A draw two lines
respectively perpendicular to the two planes, in accordance
with the theorem stated in Prob. 201.
Find the angle between these two lines by Prob. 215.
Then this angle, subtracted from 180, will give the angle
between the planes.
This solution is based on a proposition of pure solid
geometry, which states that the angle between two planes is
equal to the supplement of the angle between any two lines
respectively perpendicular to the planes. Refer to Prob.
227 for another method of solution.
222. Problem. To find the angle between a given
straight line and a given oblique plane.
One method of determining this angle has already been
given in Prob. 208. The following is an alternative
solution :
Take any point A in the given line, and through A draw
a line perpendicular to the given plane. Find the angle
between these two lines by Prob. 215. Then this angle,
subtracted from 90, gives the angle required.
This solution is based on a proposition of pure solid
geometry, which states that the angle between a line and a
plane is equal to the complement of the angle between the
line and a line perpendicular to the plane.
223. Problem. To detennine a plane which shall
contain a given straight line, and be perpendicular to a
given plane.
Take any points in the given line, and through A draw
a line perpendicular to the given plane (Art. 201). Then
the plane through the two intersecting lines, determined as
in Prob. 215, is the one required.
This construction is based on Theorem 6 of the Ap-
pendix, and is often required.
250 PRACTICAL SOLID GEOMETRY chap.
224. Problem. To determine the plan and elevation
of the line of intersection of two given planes.
Let VTH and LMN be the given planes ; four cases
are shown in the figure.
In each case let the horizontal traces intersect at r, and the
vertical traces at/, so that RS is that part of the intersection
intercepted by the planes of projection.
The elevation of R is r, and the plan of *S" is s, r and s
being in xy. Hence rs is the plan, and r s the elevation
of the required intersection.
In (b) the vertical traces intersect below xy, and RS is
between the planes forming the fourth dihedral angle. In
(d) the horizontal traces are parallel to each other, and RSis a
horizontal line parallel to either trace.
225. Problem. To determine the plan and elevation
of the point of intersection of three given planes.
The point required is that in which the line of inter-
section of any two of the planes intersects the third plane.
Or it may be determined as follows :
Find RS and PQ, the lines of intersection of any two of
the three pairs of planes. These lines, produced if necessary,
will intersect in the required point.
The accuracy of the work may be tested by observing
whether the apparent intersections, in plan and elevation, lie
on the same projector.
Examples. 1. Take two points t, m in xy 3" apart. Above xy
construct the triangle tlm, where // is 3j", and Im is 1.8".
Below xy construct the triangle t?im, making tn equal 1 . 3", and
nm 2\ ". Determine the projections of LN, the line of inter-
section of the planes L TN and LMN.
Find the inclinations of LN to the planes of projection.
2. Two vertical planes meet xy at points 2" apart, and the planes
make angles of 55 and 40 respectively with xy. Draw the
plan and elevation of their intersection.
3. An inclined plane makes 45 with the ground. An oblique
plane is parallel to xy and is inclined at 50" to the ground.
Determine the projections of the intersection of these two
planes.
THE OBLIQUE PLANE
251
4. Take any point in the intersection of Ex. 3 and draw the pro-
jections of a line passing through this point, lying in the oblique
plane, and making an angle of 55 with the intersection.
Hint. Rabat the oblique plane about its horizontal trace.
5. Draw the projections of a line AB inclined at 35 and 50
respectively to the horizontal and vertical planes. Draw the
traces of any pair of planes which intersect in AB.
6. Draw the traces of any three planes and determine the projec-
tions of their point of intersection.
252 PRACTICAL SOLID GEOMETRY chap.
226. Examples on Problems 210 to 225.
*1. In Fig. (a) the plan of a point A lying in the plane VTH is
given ; find the elevation of the point.
*2. In Fig. (/>) the projections of a point and the horizontal trace of
a plane are given. Find the vertical trace.
*3. Find the traces of the three planes which pass through A, Fig.
(e), and are parallel to the three given planes.
*4. In Fig. (d) the point whose elevation is given lies in the given
plane ; draw its plan and obtain the two rabatments of the
point. Work the corresponding problems having reference to
Figs, (a) and (/>).
*5. Find the angles between the traces of each of the three planes
of Fig. (e).
*6. Find the traces of the plane determined by the two intersecting
lines, Fig. {c). Find the angle BAC.
*7. Find the positions of B and C on the given lines, Fig. {c), so
that ABC shall be an isosceles triangle having AB = AC.= 2. 5".
*8. Determine the projections of the bisector of the angle BAC in
Fig- (<")
*9. In Fig. (e) find each of the angles which AB makes with xy,
with the trace LM, and with the trace MN.
*10. Find the angles between the planes LMN and PQN in Fig.
(e). Find also the angles between LMN and I 'TILT, and
between PQN and VTHT.
*11. Find the angle between the line AB and the plane LMN in
Fig. (e). Also between AB and PQN. And between AB
and VTHT.
*12. Determine the three planes which contain the line AB, Fig. (e),
and are respectively perpendicular to the planes LMN, PQN,
VTHT.
*13. In Fig. (e) determine the line of intersection of the planes LMN
and PQN. Also of the planes LMN, VTHT. And of the
planes PQN, VTHT.
*14. Find the point common to the three planes of Fig. (e).
15. The traces /// and vt of a plane make 55 and 45 with xy. A
point A is in the plane and ij" above the ground. Is this
information sufficient to define the position of A ? If not, what
is the locus of A? Draw the projections of the locus of A.
16. In Ex. 15 determine the positions of A, if in addition to the
data there given, you are told (1) that the point is ii" from the
vertical plane, or (2) 2" from xy, or (3) that the point is 3"
from the point t where the plane cuts the ground line.
17. Determine the intersections of the three planes of Fig. (c) with
plane which contains xy and the point A.
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^53
254 PRACTICAL SOLID GEOMETRY chap.
227. Problem. A given irregular triangular pyramid
rests with one face on the ground ; to determine its six
dihedral angles, and the true shapes of its faces.
Let abed be the given plan of the pyramid, the face
ABC being on the ground, and the height of D being as
indexed.
First to find the dihedral angle between the faces which
intersect in AD.
Take xy parallel to ad, and on xy draw ad', the
elevation of AD. Draw any line vt perpendicular to a'd\
and /// perpendicular to ad; these lines are the traces
of a plane perpendicular to AD. This plane intersects
AD in P, and the two faces ADB, ADC in PR, PS,
which two lines are perpendicular to AD {Theorem 16,
Appendix), and therefore contain an angle which measures
the angle between the faces in question {Definition 9,
Appendix).
The true angle between these lines is the angle rP s,
obtained by a rabatment of the plane VTH about th.
The angles between the other two pairs of sloping faces
may be found in a similar manner.
Next to find the angle between the faces which
intersect in PC, and the true shape of the face DBC.
Rabat DBC into the ground about be. The con-
struction for this is shown in the figure ; it is the same
as that explained in Prob. 186. The angle between the
faces which intersect in BC is measured by the angle
d 6 md", and the true shape of the face DBC is the triangle
D be.
The remaining dihedral angles, and the true shapes of
the remaining faces, may be similarly found.
Example. Draw a triangle ABC, making AB 3^", AC 2|", and
CB 3|"; take a point D in the triangle if" from A and if"
from C. Join AD, BD, CD. This is the plan of an irre-
gular pyramid with the face ABC on the ground, and the
vertex D if" above the ground ; determine the six dihedral
angles and the true shapes of its faces.
x THE OBLIQUE PLANE 255
-?Po
228. Some tangential properties of cones.
Place a cone with its base on a horizontal plane, and let
a cardboard model of a plane, having one edge on the
horizontal plane, rest against, or be tangential to, the surface
of the cone. It will be observed :
(1) That the tangent plane (if sufficiently extended)
contains the vertex of the cone ; (2) that the inclination of
the plane is equal to the base angle of the cone; and (3)
that the trace of the plane touches the circular base, or trace,
of the cone.
If the cone have its base in the vertical plane, similar
remarks apply, having reference to the vertical plane.
A plane can in general be found tangential to two cones.
(1) When the cones have their axes parallel, and their
vertical angles equal ; (2) when the cones have a common
vertex ; (3) when the cones circumscribe the same sphere.
If the student finds difficulty in the application of these
principles to any of the remaining problems of this chapter,
let him return to the problems after reading Chapters XIII.
and XIV., where the principles are illustrated in more
detail. See also the theorems in the Appendix.
256 PRACTICAL SOLID GEOMETRY chap.
229. Problem. To determine the inclinations e and
9 of a given oblique plane to the horizontal and vertical
planes of projection.
Method i. Suppose a cone with its base angle equal to
6, to be cut into two halves by a plane containing the axis ;
then one of these half-cones may be placed with its
triangular face in the vertical plane, and the semicircular
base on the ground, so that the given plane shall be
tangential to it. The following solution consists in
drawing the plan and elevation of the half- cone in this
position, and so finding the base angle 0.
Take for the vertex of the half-cone any point S in the
vertical trace, s, s being its projections. With centre .? describe
the semicircle man touching th at o. This is the plan of
the half-cone. Join / to m and n. Then tns'n is the
elevation of the half-cone, and the base angle marked 6 is
the required inclination of the plane to the ground.
To find <f>, take for the vertex of a half-cone any point R in
the horizontal trace, r, r being its projections. With centre
r, describe the semicircle pk'q, touching tv at k' . This
semicircle is the elevation of a half-cone with its base in the
vertical plane, and to which the plane is tangential. The
plan of the half-cone is rpq, and the base angle, marked c/>,
is the required inclination of the plane to the vertical plane.
Method 2, The inclination 6 may also be found by
drawing an edge elevation of the plane as in Prob. 198.
The inclination </> may similarly be obtained by drawing an
edge plan of the plane on x x y v taken perpendicular to the
vertical trace.
The model illustrating the first of these constructions will also serve
to illustrate the second, by turning it into a position so as to reverse
the planes of projection. And the construction given for the first will
also serve to illustrate the second, if it be read with the page upside
down, so as to reverse the traces of the plane.
Method 3. The construction for the rabatments of a
point A, in Prob. 213, also gives the inclinations of the
plane, the angle 6 being equal to the angle ama" of Fig.
213, and the angle 4> equal to ana l of the same figure.
THE OBLIQUE PLANE
257
&V
230. Problem. Having given one trace of a plane, and
also the inclination of the plane to one of the planes of
projection, to find the other trace.
Let the horizontal trace th be given, and also 8, the
inclination of the plane to the ground.
Take any point j - in xy, and with centre .f describe the semi-
circle tfion, touching the given horizontal trace at 0. Through
m and u draw lines making with xy, and intersecting
at s\ Then the line ts' is the required vertical trace.
Examples. 1. Determine the inclinations 6 and $ of a plane whose
horizontal and vertical traces make 30 and 50 with xy.
2. A line intersecting xy at 50 represents both traces of an
oblique plane. Find 9 and <p, the inclinations of the plane
to the horizontal and vertical planes.
3. The real angle between the traces of a plane inclined at 50 is
8o c ; represent the plane by its traces.
4. A line making 40 with xy is the horizontal trace of a plane
inclined at 50 to the horizontal plane ; determine its vertical
trace.
5. A line making 45 with xy is the vertical trace of a plane inclined
at 60 to the ground. Determine the horizontal trace.
6. A line making 6o with xy is the horizontal trace of a plane
inclined at 65 to the vertical plane. Determine its vertical
trace.
7. A line making 40 with xy is the vertical trace of a plane
inclined at 6o to the vertical plane. Determine its horizontal
trace.
S
258 PRACTICAL SOLID GEOMETRY chap.
231. Problem. To determine the traces of a plane
which shall be inclined at given angles e and 9 to the
horizontal and vertical planes of projection.
First observe, as stated in Art. 228, that if there be two
cones such that the vertex of the second cone does not fall
within the first cone, we can conceive a plane which touches
the first cone and passes through the vertex of the second.
Now this plane will also touch the second cone if the two
cones circumscribe the same sphere. See the theorems of the
Appendix, relating to the sphere, cone, and cylinder.
With any centre c in xy, and any radius, describe a
circle. This is both plan and elevation of a sphere with
its centre in xy.
Draw tangents v'r , v's to this circle as shown, each in-
clined at to xy ; and with centre c describe the semi-
circle r'as'. These are the projections of an upright half-
cone, base angle 0, circumscribing the sphere.
Also draw tangents v x e, v x f\o the circle, each making <f>
with xy, and draw the semicircle ea^f as shown. These
are the projections of a second half-cone circumscribing the
sphere, with its base in the vertical plane, and base angle
equal to </>.
Finally, from v' draw the tangent v'a^ to the semicircle
ea-[f\ and from v x draw the tangent v x a to the semicircle
r'as' . These tangents will be found to intersect in a point
/ in xy, and are the traces of the plane required.
The projections of the two generators A V, A x V v and the
point P in which the plane touches the two cones and
sphere are shown in the figure.
Limits to the data. A tangent could not be drawn from v to the
semicircle on ef, if v' were inside this semicircle ; therefore, cv' must
not be less than ce. Now, observing that cov' and co x e are right-angled
triangles, having the sides co, co. equal, it appears that if cv is greater
than ce, the angle ov'c opposite oc must be less than <p opposite o-^c.
But 6 + ov'c = go, therefore d + (p cannot be less than 90. Since
neither nor (p can be greater than 90 , d + (p cannot be greater than
180 .
Thus 6 + <P must lie between 90 and 180 , both inclusive.
THE OBLIQUE PLANE
259
Examples. 1. Determine the traces of a plane inclined at 6o
to the ground, and 45 to the vertical plane.
2. A point A is 2j" above the ground, and i|" in front of the
vertical plane ; determine the traces of a plane passing through
A and inclined at 40 and 55 to the horizontal and vertical
planes of projection.
3. Draw the traces of a plane which is inclined at 45 and 50 to
the horizontal and vertical planes of projections, the angle
between the traces being obtuse.
Hint. Take the vertical cone in the above figure, with its
apex downwards.
4. Determine two planes, one of which is inclined at 40 and 50
to the horizontal and vertical planes of projection, and the other
at 90 and 90 . Find the angle between these planes. Ans.
90.
5. Work Prob. 231 when = 6o, and ^=45, taking the radius
of the sphere to be 1". Determine the contact of the plane
with each cone and with the sphere.
260 PRACTICAL SOLID GEOMETRY chap.
232. Problem. -To determine a plane which shall be
inclined at a given angle e, and shall contain a given
straight line AB.
Find Hand V, the horizontal and vertical traces of the
line. These are points in the traces of the required plane.
Next represent a cone with its base on the ground, its
vertex at any point in the line, say A, and having a base
angle equal to 6. The required plane will be tangential to
this cone, and the horizontal trace must touch the plan
of the circular base (Art. 228).
The isosceles triangle with its vertex at a, and its
base angle equal to 0, is the elevation of this cone ; the
plan is the circle with centre a.
The horizontal trace of the required plane is the line
ht drawn through h to touch this circle, and the vertical
trace is the line drawn through / and v', where v is the
elevation of the vertical trace of BA.
233. Problem. To determine the traces of a plane
which shall be perpendicular to a given oblique plane
VTH, pass through a given point A, and have a given
inclination e.
The required plane will be perpendicular to the plane
VTH, if it contain any line perpendicular to the plane
VTH
Therefore draw ar perpendicular to ht, and dr perpen-
dicular to vt ; these are the projections of a line AR
perpendicular to the plane vth. The problem therefore
reduces to the last one.
Therefore draw the projections of an upright cone
with vertex A and base angle 9. Obtain n, the horizontal
trace of AR ; and from n draw a tangent (there are two)
to the plan of the base of the cone, thus obtaining mn,
the horizontal trace of a plane satisfying the required
conditions. The vertical trace ml is readily obtained
since A is a point on the required plane.
For examples see page 263.
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261
262 PRACTICAL SOLID GEOMETRY chap.
234. Problem. To determine the traces of a plane
passing through a given point A, having a given inclina-
tion o, and making an angle a with a given plane VTH.
If there be two cones having ias a common vertex,
their base angles being and a respectively, then if these
cones have their bases on the ground and the given plane
respectively, a plane which touches both cones will satisfy
the given conditions.
Convert the given oblique plane VTH into the inclined
plane VOH as in Prob. 198, and obtain a", the new
elevation of A on x'y.
Now draw d'r" s" and a"u"w", the elevations (on x'y) of
the cones described above.
The horizontal trace of one cone is the circle with
centre <?, and that of the other is an ellipse whose elevation
is ef.
We shall avoid having to draw this ellipse in plan, by
the following artifice :
Any convenient sphere is conceived as inscribed in the
cone whose elevation is a"e"f" ; c" is the elevation of the
centre of such a sphere. This sphere is then conceived as
circumscribed by a cone with its base on the ground, and
base angle 6 ; the plan and elevation of this cone are
shown. Observe that the plan c of the centre of the base
of this last cone will be in the line through a at right angles
to ///.
Draw a common tangent nm to the plans of the two
upright cones. This is the horizontal trace of one plane
satisfying the given conditions, and the vertical trace 1m
is readily obtained as shown, since A is in the plane.
For, a plane which touches the two vertical cones has
the given inclination ; it also passes through the given point
A, and touches the sphere centre C, and therefore is
tangential to the cone whose elevation is a"u"w".
Note 1. The trace mn might have been obtained by first draw-
ing the elliptical trace referred to above, and then drawing
a common tangent to this ellipse, and the circle centre a.
THE OBLIQUE PLANE
263
3.
4.
Examples. 1. A point A is 2" in front of the vertical plane, and
I A" above the ground. A point B is 1^" to the left of A, \"
below it, and J" farther from the vertical plane. Determine the
traces of two planes each containing AB and inclined at 50 to
the ground.
2. The traces vt, th of a plane make angles of 50 and 45
with xy. A point A is iA/' to the right of /, 2" above the
ground, and iA" in front of the vertical plane. Determine the
traces of a plane which passes through A, is inclined at 65
to the ground, and is perpendicular to the given plane.
Determine a plane inclined at 65 , passing through the point
A, and making an angle of 6o with the plane in Ex. 2.
A plane parallel to xy is inclined at 6o. Determine a second
plane which shall be inclined at 45 both to the first plane and
to the vertical plane.
264 PRACTICAL SOLID GEOMETRY chap.
235. Problem. To determine the traces of a plane
which shall pass through a given point A, make a given
angle a with a given line AB, and have a given inclina-
tion e.
First observe that the required plane will touch each of
two cones, namely :
(1) An upright cone with vertex A, and base angle 9.
(2) A cone with vertex A, axis AB, and semi-vertical
angle a.
The projections of the first cone are easily drawn, and
those of the latter may be obtained in the following
manner :
Suppose AB to turn about the vertical line Aa until it
is parallel to the vertical plane. The plan of AB will now
be aB v and the elevation a'B{.
Through a draw two lines each making an angle a with
a'B^, and describe a circle to touch these lines, the centre
being any point c^ on a B^.
Now describe a circle with the same radius, but with c
as centre, qV being parallel to xy. Draw tangents to this
circle from a, then these form the elevation of a cone of
indefinite length, having AB for its axis and a semi-vertical
angle a, C being the centre of an inscribed sphere. The
outline of the plan of this cone is not required.
Next draw the projections of an upright cone with base
angle 9 circumscribing the sphere, centre C, its base being
on the ground.
Draw nm to touch the plans of the bases of the two ver-
tical cones, then nm is the horizontal trace of the required
plane (there are two such), and the vertical trace is readily
drawn because A is a point in the plane.
It will be obvious that since the plane LMN touches
the two upright cones, it must pass through A and touch
the sphere centre C ; hence it must touch the inclined
cone.
Note. In connection with this problem the student should study
the theorems on the cone and cylinder given in the Appendix.
THE OBLIQUE PLANE
26:
Examples. 1. A line AB is parallel to the vertical plane and
distant i|" therefrom ; its inclination to the ground is 50 . De-
termine the traces of a plane which makes 30 with AB and has
an inclination of 6o.
Note. Observe that there are limits to the data in this
problem. Thus the inclinations of the line and plane being
50 and 6o, show that the angle between the line and plane
may not be any angle, but must be between o and 70.
2. Find the angle between a face of a cube and a diagonal of the
solid. Then determine a plane which contains the face and is
inclined at 65 , the diagonal of the cube being parallel to the
vertical plane and inclined at 45 .
266 PRACTICAL SOLID GEOMETRY chap.
236. Problem. The projections of a line AB being
given, and the horizontal trace, ht, of a plane which passes
through A, it is required to determine the projections of a
line AP lying in the plane HT and making a given angle
6 with the given line.
Note. The given angle must not be less than that between the
given line and plane.
First determine and set out the true length of AB.
Draw AP so that the angle BAP is [3, and choose any point
C on AP.
We can now find the projections of C because it lies on
the surface of
i. A sphere, centre A, radius A C.
2. A sphere, centre B, radius BC.
3. The given plane.
Draw x'y at right angles to ht ; project a", then oa" is
the edge view of the given plane. Project b" .
With a as centre and AB as radius describe a circle.
With b" as centre and BC as radius describe another circle.
These are the elevations, on x'y', of the two spheres men-
tioned above, and they are intersected by the given plane
in circles the elevations of whose diameters are s"w" and
r u .
Rabat the given plane and these two circles into the
ground as shown ; the rabatments of the circles intersect in
C (there are two such points).
From C obtain C ' and c"; from C" and c" obtain the
plan c, then ac is the plane of one line fulfilling the required
conditions.
The elevation of AC on xy can be drawn since the
height of C is known.
We have given the above method because the device of
employing spheres to fix the position of a point is one
which may be used in solving many problems.
THE OBLIQUE PLANE
267
Examples. 1. A line AB is parallel to the vertical plane and
inclined at 6o to the ground. An inclined plane passes through
A and makes 70 with the ground. Determine the projections
of a line AC 2" long which lies in the given plane and makes
65 with AB.
Would the solution be possible if 45 were substituted for
65 ? Ans. No.
2. A line AB makes 30" and 50 with horizontal and vertical
planes; draw its projection if.-/ is ih" from each plane. A
plane contains A and makes 45 and"55 with the horizontal
and vertical planes ; draw its traces. Determine the projections
of a line AC 2" long which lies in the plane and makes 60
with AB.
3. The traces vt, th of a plane make 50 and 35 with Ay. Deter-
mine a line which passes through 7\ lies in the plane VTH,
and makes an angle of 65 with the ground line xy.
268 PRACTICAL SOLID GEOMETRY chap.
237. Problem. An oblique plane VTH is given, and the
plan of a straight line CD which lies in this plane. It is
required to determine a plane which contains CD and
makes an angle a with the given plane.
Convert the oblique plane VTHmto the inclined plane
V'OH by Prob. 198, and project d" from d.
Draw d"e" perpendicular to ov' . Draw the elevation,
d"a"r", of a cone which has DE for its axis, and semi-verti-
cal angle equal to 90 - a.
By Prob. 274 determine aa v bb v f,f v the plans of the
axes and foci of the ellipse (but not the curve) in which this
cone intersects the horizontal plane.
Now the required plane must touch this cone, hence its
horizontal trace must touch the ellipse; it must also pass
through c, the horizontal trace of CD.
Therefore by Prob. 94 draw a tangent from c to the
ellipse ; this is ;/;;/. And ml drawn through d' will be the
required vertical trace.
Thus Imn represents the required plane.
Example. The traces vt, th of a plane make angles of 45 and
35" with xy. C and D are points in the traces, each distant
2.5" from t. Determine a plane which contains CD and makes
an angle of 50 with the plane vth.
238. Examples on Problems 225 to 237.
1. A point is ij" from xy. State and project its locus. Ans. A
cylinder of indefinite length. The projections are straight lines
distant Ij" from, and parallel to xy,
2. A point is 1^" distant from a point c in xy. State and project
its locus. Ans. A sphere 3" diameter with centre at c. The
projections are coincident circles.
3. A point is 1" above the ground. State and project its locus.
Ans. A horizontal plane 1" high. The vertical trace is parallel
to xy.
4. A point is 1 J" from xy, and 1^" distant from the point c in xy.
State and project its locus. Ans. The circle in which the
cylinder and sphere of Exs. I and 2 intersect. Projections,
coincident straight lines perpendicular to xy.
5. A point is Ij" from xy and 1" above the ground. State and
project its locus. Ans. The two horizontal lines in which the
cylinder of Ex. 1 is cut by the plane of Ex. 3.
X
TFIE OBLIQUE PLANE
269
6. A point is distant l|" from the point c in xy, and 1" above the
ground. State and project its locus. Ans. The horizontal
circle in which the sphere of Ex. 2 is cut by the plane of Ex. 3.
7. A point is I J" from xy, ij" from a point c in xy, and 1" above
the ground. Determine its position. Ans. Either of the two
points common to all the loci of Exs. 1 to 6.
8. The traces of th and tv of a plane make 55 and 45 with xy. A
point A is 2" distant from t, i\" from the plane VTH, and
1 y from the vertical plane. Find all the positions of A which
satisfy these conditions.
9. Three planes are mutually perpendicular. One is inclined at
40 , a second at 6o ; find the inclination of the third.
10. Three lines are mutually perpendicular. One is inclined at 40 ,
a second at 30 ; find the inclination of the third.
11. Draw the complete plan of an equilateral triangle ABC of 3"
edge, having given the plan ab is 2\" long and the plan ac
makes 30 with ab.
270 PRACTICAL SOLID GEOMETRY chap.
239. Miscellaneous Examples.
*1. Determine the true angle between the lines AB and BC.
*2. Determine the projections of a circle of l|" diameter lying in the
plane containing the two given lines AB, BC, produced if
necessary, and touching both. ('891)
*3. The traces of a plane are given and the projections of a point P.
From P draw two lines each 2j" long, meeting the plane in
points on the same horizontal line and 2^" apart. (1890)
*4. From the given points A, B, draw two lines meeting on the
given plane and making equal angles with it. (1891)
*5. A /' are the projections of a given point P ; ab, a'b' those of a
given line AB. Draw the projections of an equilateral triangle,
with one vertex at P and the side opposite that vertex on the
line AB. (1897)
*6. a is the plan of a point A lying in the given plane VTH.
Through A draw a line in the given plane VTH parallel to the
given plane BOC. (1888)
" *7. Draw a plane perpendicular to the plane of the triangle ABC
and bisecting the sides BC and AB. ('894)
8. Draw the traces of any plane equally inclined to the planes of
projection, and determine the true angle between its traces.
9. A plane is equally inclined to both co-ordinate planes, and the
real angle between the traces is 50. Draw the traces. (1892)
10. The vertical trace of a plane makes 40 with the ground line,
and the plane is inclined at 50 to the horizontal plane. Draw
its horizontal trace. Determine the point (P) in the plane 1"
in front of the vertical plane and I h" above the horizontal plane,
and through P draw a line in the plane making equal angles
with its traces. (1889)
11. Draw a plane inclined at 45 to the horizontal plane, and at 6o
to the vertical plane. Draw a line in the plane inclined at 30 ,
and 2" long between its traces. Lastly, draw the plan of a
regular hexagon lying in the plane of which the above line is a
diagonal. ( 1897)
12. Draw the traces of any plane inclined at 40 . Determine the
projections of a line in this plane inclined at 27 . Draw the
traces of a plane containing this line, and inclined 63 . Find
the angle contained by these two planes. Determine also the
perpendicular distance between xy and the intersection of the
two planes.
13. The traces vt, th of an oblique plane make 30 and 50 with
xy. Take a point a 1" from ht and 2" from xy ; this is the
plan of a point A 2^" above the ground. Draw af 2J" long
making 15 with xy. Determine the height of F (1) if AF
is 3-n" long, and F is lower than A ; (2) if AF meets xy.
THE OBLIQUE PLANE
271
Cofiy /tie figures double size. 7? >
272 PRACTICAL SOLID GEOMETRY chaf.
*14. ad is the plan of a line in a plane whose traces are ab, be. One
side of an equilateral triangle of 1.5" side lies along ad, one
extremity at a. The plane of the triangle is inclined at 40 to
the horizontal plane. Draw the projections of the triangle.
(1895)
" ;f 15. A line AB and a plane are given. Through AB draw a plane
perpendicular to the given plane and determine the line bisect-
ing the angle between AB and the line in which the planes in-
tersect. (1892)
*16. ABC is the base of a tetrahedron, and is in the horizontal
plane. D is at 2.5" above it. (a) Find the values of the
dihedral and plane angles round the vertex D. (b) Pis a point
in the base ; find the points where perpendiculars from P meet
the three other faces of the tetrahedron. ^895)
*17. Determine the traces of any two planes containing the given
line AB and including an angle of 6o. (1884)
*18. / is the plan of a point P distant h" (measured perpendicularly
to the surface) from the plane VTH and above it. Through
P draw
(1) a line parallel to the plane VTH, and inclined at 45 to
the horizontal plane ;
(2) a line also parallel to the plane VTH, but making an angle
of I 5 with the line inclined at 45. (1896)
*19. AB is a given line, C a given point. Find the scale of slope of
the plane of A, B, and C ; and draw the plan of a square in that
plane, one diagonal to be a perpendicular let fall from C on the
line .-^9. Unit = o.i". (1895)
*20. Determine a line through the point C to meet the line AB at a
point D such that the angle CDA shall be equal to 50 .
Unit = 0.1". (1894)
21. Draw an arc of a circle 3" in diameter with centre v. Along
the circumference set off adjacent chords ab, be, 2" and 1^"
long. Join av, bv, cv. Assuming the circle to be in the hori-
zontal plane and the height of V to be 3" above this plane, de-
termine the angle made by the plane ABV, with the plane
CBV. (1893)
*22. Draw (1) a plane (M) inclined at 65 to the horizontal plane,
and containing AB ; (2) a plane (7V) through B inclined at
50 to the horizontal plane, and perpendicular to M. (1896)
23. The horizontal and vertical traces of a plane make angles of 35
and 40 with xy. Draw the plan of any line lying in this
plane and inclined 30 . Now draw the plan of a line which is
parallel to this, and lies in the plane, so that the part of it lying
between the traces of the plane is 2".
THE OBLIQUE PLANE
273
Copy the figures double size.
o
12 -5
CHAPTER XI
HORIZONTAL PROJECTION, OR FIGURED PLANS
240. Problem. Having given the indexed plans of
three points A, B, C, to determine the indexed plan of a
horizontal line which passes through A and intersects BC.
Case I. Let a s , b vi , c 6 be the given plans.
First Method, Fig. (i). Join b 12 c 6 , and draw b v2 b' and
c t .c perpendicular to b l2 c 6 , and equal to 12 and 6 units
respectively 5 join b'c .
Make b V2 h equal to 8 units, and draw /id', d'd respect-
ively parallel and perpendicular to b r / G ; affix the suffix
8 to the point d, then a s d $ is the required plan.
Second Method, Fig. (2). Draw any xy which is not
perpendicular to b v> e 6 , nor so nearly perpendicular as to
lead to an ill-conditioned construction, and obtain a, b', c
the elevations of A, B, C.
Join b'c, and draw ad' parallel to xy ; draw the projector
d'd; affix the suffix 8 to d ; then a s d 8 is the required plan.
Note. By this method xy may be drawn with the tee-square, and
the set-square used for the projectors.
Case II. Let a v Ik, c b be the given plans, Fig. (3).
The line BC is nearly horizontal, and the point on it at the
level of A is supposed to be inaccessible.
Determine the elevations of A, B, C as in the second
method of Case I. Join a'b' ; draw e'e parallel to xy ; pro-
ject / to e 5 ; join e 5 c 5 , and draw a Y d x parallel to e 5 c 5 . Then
a x d x is the required plan.
xi HORIZONTAL PROJECTION, OR FIGURED PLANS 275
Unit = 075"
241. Problem. To determine the scale of slope of a
plane containing three points A, B, C, the indexed plans of
which are given.
Let g 8 , ?> 10 , <r 6 be the given indexed points, Fig. (2).
By one of the methods of Prob. 240 determine a^d^,
the plan of a horizontal line in the plane of ABC.
Draw any double line at right angles to d s a s . Let d & a 8
and a line through b 12 , parallel to d s a 8 , meet this double
line in gmf Then the requited scale of slope is g s f lT
Example. Draw a triangle 20 Vi5' making a%p % 2^", a 2( f yo 3", and
8 f 15 3|". Draw the plan of a horizontal line CD which inter-
sects AB. Draw an elevation of ABC on a plane at right
angles to CD. Draw a scale of slope for the plane of ABC.
276 PRACTICAL SOLID GEOMETRY chap.
242. Problem. Having given a plane by its scale of
slope ab, to determine (1) the inclination e of the plane ;
(2) a scale of slope of the plane which passes through the
given point P and is parallel to the given plane ; and (3) a
scale of slope of the plane which passes through P, is per-
pendicular to the given plane, and is inclined at a given
angle 9.
(1) Draw xy parallel to ab, and obtain the elevation a'b'
by making nd =10 units, and mb' = 30 units ; join b'd, and
produce to meet xy. Then a'b' is an edge view of the
plane, and the angle marked 9 is the required inclination.
(2) Obtain p', the elevation of P, and draw p'e parallel
to b'd ; then p'e is an edge view of the required plane, and
e'e drawn perpendicular to xy is its horizontal trace.
At any point e in this trace draw the double line parallel
to ab, and on it mark off ed equal to, say, 30 units measured
from the scale of ab ; index the two points of the scale o and
30 as shown, and the scale of slope of the required plane is
completed. Observe that since the planes represented by ab
and ed are parallel, equal differences of level of the two
planes correspond to equal lengths on their scales of
slope.
(3) Every plane passing through P at right angles to the
given plane ab must contain the line through PaX. right angles
to that plane.
Draw p'Ji perpendicular to db', and P u b Q parallel to ab ;
then these are the projections of the line through P per-
pendicular to the given plane, and H is the horizontal trace ;
hence the horizontal trace of the required plane will pass
through Ary.
Draw the projections of a cone with its base on the
ground, its base angle <, and its vertex at P ; the required
plane will touch this cone, and the horizontal trace will
touch the plan of the base of the cone.
The plan of the cone is the circle, centre p n ; hence the
horizontal trace of the required plane is the tangent h t
drawn from /i Q to this circle.
xi HORIZONTAL PROJECTION, OR FIGURED PLANS 277
Unit = .031
Take any point/ in // t ; draw the double line^ r perpen-
dicular to //,/, and draw pg parallel to h Q t ; then ft is the
horizontal trace of the required plane, and p 41 g is the plan
of a horizontal line at the level 41 ; hence the scale of slope
fg can be indexed as shown.
Examples. 1. Draw a double line ab 2" long, and attach indices
of 10 and 30 to a and b. Regard this as the scale of slope of a
plane. A point c 95 is 3" from a, and 2^ from b. Determine
the scale of slope of a plane (a) through C parallel to the given
plane ; (b) through C perpendicular to the plane and inclined
at 65 to the ground. Unit = o.i".
2. Determine the plan of a line in the plane (a) Ex. 1, at a level 15,
and of one at the same level in the plane (/') ; show the plan of
the intersection of these lines, and index it.
278 PRACTICAL SOLID GEOMETRY chap.
243. Problem. The figured plan c 25 d 13 of a line, and
the scale of slope ab of a plane are given ; to determine
(1) the figured plan of the point of intersection of the line
and plane ; (2) the angle between the line and plane ; and
(3) the bisector of this angle. Unit OT'.
(1) Take ab as a ground line, and obtain ab' the edge
view of the given plane, and c'd! the elevation of the given
line CD ; these intersect in /'. Draw the projector i'i,
then i is the plan of the point of intersection of the given
line and plane, and its index may be obtained by measuring
i' in.
(2) Determine en', e 2f n, the projections of CJV, the
perpendicular from C to the plane. Then obtain Z JV Q by
rabatment of the plane about its horizontal trace ; and next
find C by rabatment of the triangle INC about I Q N .
Then JV I Q C is the required angle between the line
and plane.
(3) Bisect the angle JV f C by I^H^. This is the
rabatment of the required bisector.
Set off n'li! equal to N^H^ and project h from ft. Then
ih is the plan of the bisector. The indices for i and h
may be found by measuring the heights of i' and ft above
ab.
Example. Draw a triangle abc, making ab 2", ac=$", and
be = 2.5". Take d 2" from b and if" from c, and outside the
triangle abc. Index the points a and d, 8 and 25 ; this is the
plan of AD. Regard be as the scale of slope of a plane, B
and C being at levels 10 and 30. Unit o. 1".
(a) Determine the indexed plan of the point of intersection
of the given line and plane.
(b) Find also the angle between the line and plane.
(c) Draw the indexed plan of the bisector of the angle {!>).
(d) Draw the scale of slope of a plane which bisects the line
AD at right angles.
(e) Determine a scale of slope of the plane which contains
AD and is parallel to BC, and of the plane which con-
tains BC and is parallel to AD. Find the distance
between these parallel planes.
xi HORIZONTAL PROJECTION, OR FIGURED PLANS 279
c
2S0 PRACTICAL SOLID GEOMETRY chap.
241 Problem. To determine the line of intersection
of two planes given by their scales of slope, ab and cd.
Case I. The method here employed depends upon the
fact that two horizontal lines at the same level, and one
on each plane, must intersect each other at a point which
lies in both planes, that is, on their intersection.
Through the division o on each scale draw lines aj, cj
respectively perpendicular to the scales of slope ; these are
the plans of two horizontal lines at a level o, one on each
plane, hence j is the indexed plan of a point on the line of
intersection. Similarly, bi and dl drawn through the divi-
sions 10 determine z 10 , the plan of another point on the line
of intersection ; hence If is the required intersection.
Case II. Let the scales of slope be nearly parallel ;
the above method will be inconvenient.
Conceive both planes to be cut by any vertical plane the
plan of which is, say, /;;/. Draw the plans of the horizontal
lines through a and b on one of the given planes, and
through c and d on the other, intersecting hn in g, h, e',f
respectively. Obtain g'ti and e'f , the elevations of GHand
EF, on //;/ as ground line ; they intersect in i, from which
the plan i, in /;//, is determined by projecting from i' . Now
the point / is in the plane ab, since it is in the line GH
contained by this plane ; and / is in the plane cd, since it
is in IE; therefore lis in both planes, or is one point in
their required intersection. On measuring W the index for
i is seen to be 7.5.
By taking any other vertical plane, say no, the point j 8
is obtained in a similar manner ; hence / 7 . 5 / s is the required
indexed plan.
This method is applicable to any case where the line of
intersection is within the limits of the paper.
Case III. (no figure). Let the scales of slope be
parallel to each other. In this case the horizontal lines
on the two planes are parallel to each other, and the line
of intersection of the planes is horizontal. We may proceed
as in II., or as follows :
xi HORIZONTAL PROJECTION, OR FIGURED PLANS 281
Take xy parallel to ab or cd, and obtain a'b\ c'd', the edge
views of the two planes ; let these intersect in /'. Then i' is
the end view of the line of intersection, and the plan ij may
be obtained by a projector from 1. The distance of i' from
xy determines the indices of i and j.
Example. Draw a quadrilateral abed, making ab = bc=z", ad=
2tt", abc=jo, bad =6o. Regard a. M d and b. l; c w as the scales
of slope of two planes ; determine the indexed plan of their
line of inteisection. Unit 0. 1".
282 PRACTICAL SOLID GEOMETRY chap.
245. Problem. To determine the scale of slope of the
plane which bisects at right angles the angle between two
lines AB, AC, whose indexed plans are given.
The required plane must contain the line which bisects
the angle BAC. It must also contain the line through
A at right angles to the Diane of BAC. These two lines
define the plane, and their projections are first found.
On b 30 a w produced determine n v the plan of a point N
on the same level as c, by the method of Prob. 240 ;
join c x n v Then CN is a horizontal line in the plane of
ABC.
Draw xy at right angles to cn x , and obtain cab', the
edge elevation of the triangle CAB. Now conceive that
the triangle revolves about CN until it is horizontal ; its
plan while in this position may be obtained thus :
Through c draw w'z' parallel to xy, and with c as
centre describe the arcs b'B ' and a'A ' ; draw bB and
and aA parallel to xy to meet projectors from B ' and A Q '
in B and A . Then A B c i is the required plan.
Bisect the angle B A c i by the line A D . Find d', the
elevation of D when the triangle ABC is brought back
to its original position ; also, from d' project d 17 on b S0 c 4 ,
measuring the index 1 7 from the elevation.
Through a draw a'h' perpendicular to cab' ; this is the
elevation of the line at right angles to the plane of the
triangle, and a 10 /i Q drawn at right angles to n^ is its plan.
The required plane contains AD and AH, and its
horizontal trace could be determined by obtaining the
horizontal traces of AD and AH, and then joining these ;
the scale of slope would be at right angles to the joining
line, and could be indexed, since the height of a point A in
the plane is known.
In the figure the horizontal trace of AD is not found.
We avail ourselves of the fact that a line which inter-
sects AD and AH must lie in the required plane.
Draw d'e parallel to xy ; this is the elevation of a hori-
zontal line intersecting AD and AH; its plan d l7 e l7 is
:i HORIZONTAL PROJECTION, OR FIGURED PLANS 283
obtained by projecting from d' and e on to b m c< and h Q a 10 ,
and measuring and indexing the level 17.
Draw the scale of slope at right angles to e^d^, meeting
the latter in s, and draw h Q r parallel to e 17 d ir to meet the
scale of slope in r. Then rs, with the indices o and 1 7, is
the required scale of slope.
Example. Determine the scale of slope of the plane which bisects
at right angles the angle BAC in the example page 275. Also
that for the angle ABC.
284 TRACTICAL SOLID GEOMETRY chap.
246. Problem. To determine the angle between two
planes given by their scales of slope ab and cd.
Determine L J the plan of the intersection of the planes,
Prob. 244.
Let a plane be taken at right angles to the intersection,
cutting the given planes in two lines and the horizontal
plane in a third line ; these lines form a triangle with its
base on the ground, the vertical angle being the one
required.
Obtain i'j the elevation of the line of intersection, on
/' j as ground line. Draw vt, th the traces of an inclined
plane at right angles to IJ. Let vt cut j Q t" in r '. Draw
/ a, j c perpendicular to ab, cd, and produce these lines to
meet h Q t in h , f .
Now rabat the triangle FRH about its base ; that is,
with centre /, draw the arc r R w and join /i R , f R - We
thus obtain f R /i Q , the required angle between the planes.
247. Problem. To determine the scale of slope of the
plane which bisects the angle between two planes given
by their scales of slope ab, cd.
Determine f R Q /i , the angle between the given planes in
the manner explained in Prob. 246, and draw ^ bisecting
the angle f R Q /i , and meeting f Q /z in g .
Now the triangle f R /i is the rabatment of the triangle
FRH about FH, and R g is the rabatment of the bisector
RG. The required plane will bisect the plane angle FRH,
and will therefore contain RG.
But since G is on FH it will not move during the rota-
tion of the triangle FRH, hence the required plane con-
tains g , which coincides with G. It also contains JI, the
intersection of the given planes.
Join Jq^q- Then since j is the horizontal trace of JI,
J0S0 wl ^ b e tne horizontal trace of the required plane.
Draw the double-lined scale of slope at right angles to
g j and meeting it in s; draw / 20 / parallel to ( ^; index s and /,
o and 20 respectively, then si is the required scale of slope.
xi HORIZONTAL PROJECTION, OR FIGURED PLANS 285
246 and 247
Examples. 1. Draw a quadrilateral abed having given sides ab
4.5", /v = 4.o", ^=2.5", da = 2.0" ; diagonal ^=4.0". -Let
the indices of the points a, b, c be 10, 30, 5. Unit = o. 1". If
this figure is the projection of a plane quadrilateral ABCD,
index the plan d.
2. Let the index of d be 40. (a) Find the angle between the two
planes of which a 1Q b i0 , c.J>. m are the scales of slope, (b) Find
the angle between the planes ABC, DEC oi the gauche quadri-
lateral ABCD.
3. Determine the scale of the slope of the plane which bisects the
angle between the planes 10 /' 40 , c : b m of Ex. 2. Also work this
problem for the planes ABC, DEC of the same example.
286 PRACTICAL SOLID GEOMETRY chap.
248. Problem. The indexed plans a 10 , b., of two points
A, B, and the scale of slope c d., of a plane are given ; it
is required to determine a point P in the given plane
which shall be distant 9 units from A and 12 units from
B. Unit = -05 ".
Since the required point P is io units from A it must
be situated on the surface of a sphere with A as centre and
radius io units. Similarly P must be situated on the
surface of a second sphere, with B as centre and 1 2 units
as radius. It is also contained by the given plane.
Hence the required point is either of the two points
where the circles in which the plane intersects the spheres
cut one another. These points are determined in the
following manner :
Take cd as a ground line and project cd' the edge
elevation of the given plane, and also d, b' the elevations
of A and B. With d as centre and radius 9 units describe
a circle intersecting cd' in u and r. With b' as centre
and radius 1 2 units describe another circle intersecting cd
in w and /. These circles are the elevations of the
spheres referred to above, and ur\ w's are the elevations
of the circles in which the given plane intersects these
spheres.
Draw dm and b'ri at right angles to cd', then M and N
are the centres of the circles.
These circles intersect each other in two points (P and
<2), each of which satisfies the required conditions ; the
projections of P will now be determined.
Conceive the plane with the two circles to be turned
into the ground about its horizontal trace. We thus
obtain the rabatments of the two circles, viz. the circles
with centres M and JV ; these intersect each other in
two points, one of which is P . Obtain p' by a process
the reverse of that by which M and JV were obtained
from m' and ri ; from p' project the plan p n , the index
being obtained by measuring the height of /'. Thus
the figured plan of the required point P is found.
xi HORIZONTAL PROJECTION, OR FIGURED PLANS 287
uy /
!
0y< -'' ;
.2*7
r i : i r i
Mil!
^
,-'' 1 \
La
! \ \ i V
^
'.-'<?
a m M i .
\ " lb l
/ |
\2f
/z<
A,
A 7 ^. This problem is of considerable importance in so far that
many other problems may be reduced to it.
The point P was determined by regarding it as one of the two
points of intersection of three surfaces two spheres and a plane. Now
there are two methods of determining such a point. We may deter-
mine the intersection with each other of any two of the surfaces, and
then obtain the points in which this intersection meets the third
surface ; or we may determine the intersections of one of the surfaces
with each of the other two, and afterwards obtain the points in which
these two intersections meet each other. The latter method has been
adopted in the above solution.
*
Examples. 1. Draw a quadrilateral (/';so f 28 </ 'i5 ma ^ n g rt (/';so = 2 >
b vf<8L = 1 5 Vis = 7 5 <V-28 = 2 > ^15 = J 7- Regard a^ as
the scale of slope of a plane, and determine the indexed
plan of a point on the plane, distant 10 units from C, and 15
units from D. Unit = 0.1".
2. In Ex. 1 determine the indexed plan of a line lying in the given
plane and making (>o with the given line.
2S8 PRACTICAL SOLID GEOMETRY cuap.
249. Problem. To determine the shortest line MM
between two given lines, AB, CD.
Let a i 1> V c\ d Q be the given indexed plans of the lines
AB, CD.
Suppose that from any point in one of the lines, say B
in AB, a line BE be drawn parallel to the other CD.
Then ABB determines a plane through AB parallel to
CD. Let now CD be projected on this plane, and let the
projection cut AB in M. At M erect a perpendicular to
the plane. This perpendicular will meet CD in the point
we have called N. Then MN\% the shortest line required.
It is perpendicular to both AB and CD.
Through b 12 draw b Y < % parallel and equal to c 10 d , the
index 2 of e being determined so that the difference of the
indices of b and e is the same as that of c and d, viz. 10.
Then BE is parallel to CD.
Determine^ and g , the traces of BA and BE; or by
Prob. 240 determine the plan of any other horizontal line
in the plane of ABE.
Draw xy perpendicular to the horizontal line ; and on
xy project the elevations f, a', b', c , and d'. Then fa'b'
is the edge elevation of the plane ABE, and c'd' will be
parallel to fab'.
Select any point in c'd', say c , and draw c'h' perpendicular
to a'b'. Project from }i to //, where c lQ h is perpendicular
to f g . Through h draw km parallel to d e 1() . Draw mn
perpendicular X.0 f^g^
Then MN is the shortest line required. The indices
of m and n may be found by projecting tri, n as shown, and
then measuring the heights of these points above xy.
Examples. 1. Two lines </' 18 , c^t^, each 2.5" long, which
bisect one another at 6o, are the indexed plans of two lines
AB, CD. Determine the indexed plan of MN, the shortest
line between them.
2. Measure the angle between AB and CD.
3. Determine a line PQ which meets AB and CD each at an angle
of 6^.
xi HORIZONTAL PROJECTION, OR FIGURED PLANS 289
P2'
4. Draw a triangle aoc, having ac
2i"
ao = 1
3"
-ll'
Regard
ac as the plan of one edge of a regular tetrahedron ; the plan
of an adjacent edge AB coincides, in direction only, with ao ;
complete the plan of the tetrahedron, the indices of a and c
being 12 and 25 respectively. Unit = o. 1".
Hint. See Prob. 248. Find the true length of an edge AC
of the tetrahedron ; then the point B is fixed in the following
way. (1) It lies on a sphere, centre A, radius AC; (2) it lies
on a sphere, centre C, radius AC; (3) it lies on a vertical
plane whose plan is given by the line ao. Therefore draw the
projections of these spheres, and find (by rabatting the vertical
plane) the circular sections of the spheres by the vertical
plane ; the circles intersect in two points either of which is
the rabatment of B. The plan b may then be readily
determined.
A line a () l>., n , 2 2 " ' on S> ^ s tne indexed plan of a square ABCD ;
unit = o. 1". The plan ad of an adjacent side makes 45 with
ab. Complete the indexed plan of the square.
Suppose ABCD of Ex. 5 to be the base of a right pyramid 3"
long, vertex V; draw the indexed plan of the pyramid.
Draw the indexed plan of an equilateral triangle ABC, having
given a d 1Q = 2", a i\ Q c=s\
U
290 PRACTICAL SOLID GEOMETRY chap.
250. Miscellaneous Examples.
*1. Determine a plane, bisecting the angle between the two given
planes. (1886)
*2. ab is a given line, CD the scale of slope of a given plane.
Determine the projection of the line ab on that plane.
Unit = o.i". (1897)
*3. Determine the intersection of the three given planes A, B, and
cod. Unit 0.1 ". (1S88)
*4. cab is the horizontal trace of a plane inclined at 40 to the
horizontal plane, cd is the plan of a line CD in that inclined
plane. Draw the traces of a plane inclined to the first plane
at 40 , the intersection of the two planes to be the line
CD. (1897)
*5. The figured plan of a triangle ABC is given ; ef is the plan of a
line which is bisected by the plane of the triangle ABC.
Obtain the index of/. (1S93)
*6. Find the common perpendicular to the two lines AB, CD, and
give its length. Unit = o.i". (1897)
*7. Two non-intersecting lines AB and CD are given (see figure
for Ex. 6). Determine the traces of a plane containing CD,
and parallel to AB, and determine the projection of AB on
this plane.
8. Two lines ab, cd, at right angles, are the plans of the centre
lines of two horizontal shafts, AB, CD, one of which is two
feet above the other. They are connected by a third shaft,
intersecting them in P and Q, such that PQ makes angles of
60 and 45 respectively with AB and CD. Draw the plan of
PQ. Scale ^ inch to the foot.
*9. Determine the angle between AB (Ex. 6) and a line AE which
intersects CD and lies in the same vertical plane as AB.
*10. AB and CD (take the figure of Ex. 6, but alter the index of b
to 16), two non-intersecting straight lines, are given by their
figured plans. Frflm A draw a line making an angle of 45
with AB and intersecting CD. Unit o. 1".
Hint. The required line must lie on the surface of a cone,
vertex A, axis AB, semi-vertical angle 45 ; it must also be in
the plane A CD. Draw the plan of the cone, and take any
vertical plane perpendicular to AB cutting the cone in a circle ;
draw the elevation (on this vertical plane) of the circle. Take
any two points on CD, say C and D. Find the points E and F
in which AC and AD meet the vertical plane, then either of
the points, in which the line EF meets the circle, when joined
to A will give a line satisfying the required conditions.
Observe that EF is the intersection of the plane A CD and
the vertical plane cutting the cone.
xi HORIZONTAL PROJECTION, OR FIGURED PLANS 291
CHAPTER XII
PLANE AND SOLID FIGURES IN GIVEN POSITIONS
251. How position is defined. Let the student take a
square cut out in paper, and place it so as to lie on a plane
surface which is inclined at a given angle, one side of the
square making another given angle with horizontal trace of
the plane ; then he will observe that the shape of the plan is
always the same no matter what position the square may
occupy on the plane, so long as the two angles remain
unaltered. The shape of the plan in this case is therefore
made definite by two angles being given. If we suppose
the square to be one face of a cube, then the shape of the
plan of the cube is also definite.
The angular position relatively to the horizontal plane
may be fixed in other ways ; for example, by having given
the inclination of two sides ; or, what amounts to the same
thing, the differences in the heights of three corners of the
square. The shape of the plan is again definite.
The shape of the elevation is not completely defined by
the conditions just stated, but depends further on the
position which the object occupies relatively to the vertical
plane of projection. If the shape of the elevation is to
be made definite as well as that of the plan, one more
angle must be given ; e.g. the angle which one side makes
with the vertical plane.
chap, xii FIGURES IN GIVEN POSITIONS 293
Reasoning thus, it is seen that the shape of the plan of a
polyhedron is definite when any one of the following sets
of conditions are given :
(a) The inclination of a face and that of a line in the
plane of the face.
(b) The inclinations of two lines, or the heights of two
points above a third, all connected with the solid,
if) The inclinations of two faces.
(d) The inclination of a face, and that of a line con-
nected with the solid, not in the plane of the face.
The shape of the elevation is also definite if, in addition,
one of the following conditions be given :
(e) The inclination of a line to the vertical plane, or the
difference of the distances of two points from the
vertical plane.
(/) 77ie inclination of a face of the solid to the vertical
plane.
It should be observed that the conditions stated in the
preceding article do not define completely the position
of a figure in space relatively to the surrounding objects,
but only its angular position. Complete definition of
position may be obtained as follows:
Let three planes mutually perpendicular be taken as
planes of reference, then the position of a point in space is
defined if we know
(g) The distance of the point from each of the three
planes of reference.
The position of a finite line is definite if we know
(//) The position of one point as in (g), and the inclina-
tions of the lifte to two of the planes of projection.
The position of a polygon or polyhedron in space is
defined if we know
(/) The position of one point as in (g) and also the
angular position relatively to the planes of refer-
ence. To define the latter three angles are necessary,
which may be those of (a) and (e) in the preceding
article, or other combinations.
294 PRACTICAL SOLID GEOMETRY chap.
252. Problem. To draw the plan of an equilateral
triangle, l 1 " side, the plane of which is inclined at e, and
one side at a.
Draw a plane inclined at 0, and in it draw any line AB
inclined at a, and obtain bA the rabatment of the line as
described in Prob. 191 and here repeated.
Draw an equilateral triangle C D Q E of the given size,
with one side parallel to, or in, bA ; let this be the rabat-
ment of the triangle about th. The plane of rabatment is
now to be turned back into its original position vth, carrying
the triangle along with it ; the projections of the triangle
will then be cde, c'd'e', obtained by the construction of
Prob. 186 reversed. The circular arcs with / as centre are
the elevations of the paths of C, D, E, the plans of these
paths being C c, -D d, E^e, perpendicular to ///.
The problem is thus solved. The plane of the triangle
is inclined at 6, and the side CD at a to the ground.
253. Problem. Draw the plan of a cube of given
edge, with one face inclined at e, and a diagonal of that
face at a.
Begin, as in the last problem, by finding B M , the
rabatment of a line inclined at a, lying in a plane inclined
at 6. The construction lines for this are not shown.
Draw the square A B Q C Z> , of the given side, with the
diagonal A C parallel to E M . Then A Q B C D may be
taken as the rabatment of the face of the cube, and also as
the plan of the cube in the rabatted position.
The plane of rabatment must now be turned back into
the original position vth, and the elevation of the face
ABCD in the plane obtained as in the last problem.
The elevation of the cube is completed by drawing the
lines a'd , b'b', c'c' , d'd' perpendicular to tv, each equal to the
edge of the cube, and then joining a'c.
The plan of the cube is now found by projecting the
points in elevation on to the lines through corresponding
points of the rabatment, drawn at right angles to th.
XII
riC.URES IN GIVEN POSITIONS
295
Examples. 1. Draw the plan of a square, .2" side, the plane of
which is inclined at 50 , one side being inclined at 35.
2. Draw the plan of a square ABCD, 2" side, when the line joining
A to the middle point of BC is inclined at 35 and the plane of
the square at 50 , the point A being on the ground.
Determine the inclinations of the diagonals of the square.
Draw the plan of a regular hexagon of ij" side in any position
such that its plane is neither horizontal nor vertical.
A regular hexagon of ij" side has one side in the horizontal
plane. The plane of the hexagon is vertical, and inclined at
43 to the vertical plane of projection. Draw the elevation of
the hexagon.
Draw the plan of a cube of 2-|" edge, one face being inclined
at 50 and one side of that face at 35.
An isosceles triangle, base 2.5", sides 3", has its base inclined at
35, and its plane at 50 ; this is the end of a right prism 2^"
long. Draw the plan and elevation of the solid.
Draw the plan of a square pyramid, side of base 2", height 3",
when the base rests on a plane inclined at 60, one diagonal of
the base making 50 with the horizontal trace of the plane of
the base.
3.
4.
5.
6.
296 PRACTICAL SOLID GEOMETRY chap.
254. Problem. An octahedron, 2" edge, has one face
resting on a plane inclined at 30 , one side of the face
making an angle of 70 with the horizontal trace of the
plane. Draw the plan and an elevation of the solid.
Let VTHhe. the plane inclined at 30 . Draw any line
Z 7l/ making an angle of 70 with the horizontal trace, and
construct an equilateral triangle A B Q C , having one side in
the line Z Af Q . Let A B C be the rabatment of that face
ABC of the octahedron which rests on the plane at 30 ,
AB being the side of the face which makes an angle of 70
with the horizontal trace. Complete the plan of the solid
in its rabatted position as follows :
Inscribe a circle in A B C , and draw f x d v d x e v e x f v
tangential to the circle, respectively parallel to A B , C Q A ,
B C . Then d x e x f x is an equilateral triangle, and is the
plan of the upper face of the octahedron. The plan of the
solid is completed by drawing the outline shown.
Now let the plane of rabatment be turned back to the
original position vth ; the elevation a'b'c of the face ABC
is readily found as in previous problems, and the construc-
tion is evident from the figure. .
To determine the elevation of the face DEF, first find
the distance between the parallel faces ABC, DEF. This
is done in the figure by finding B B> , the rabatment of
BD about its plan B d y That is, d x D is drawn at right
angles to the plan, and B D made equal to BD, \\" .
Then d x D is the distance between the faces.
To find the elevation of D determine the point s, as
shown, / being the elevation of the foot of the perpendicular
from D to the plane of the face ABC, and draw s'd' per-
pendicular to tv and equal to d x D . In a similar manner
e and/" may be found. The elevation of the octahedron
is then completed by drawing the lines representing its
edges.
The determination of the plan is left as an exercise.
Example. An octahedron 2" edge has a face inclined at 6o,
and one side of that face at 40'. Draw its plan.
XII
FIGURES IN GIVEN POSITIONS
297
Examples on Problems 255 to 258.
1. The sides of a square are inclined respectively at 30 and 45 ;
draw the plan and determine the inclinations of the diagonals.
2. Draw plan of a square pyramid, side of base I7?", height 1^",
when two sides of the base are inclined at 30 and 40 re-
spectively. Determine the inclinations of the sloping edges.
3. Determine the plan of a hexagon of 1" side, when two adjacent
sides are inclined 30 and 40 J respectively.
4. Draw the plan of a regular hexagon ABCDEF, \\" side : (a)
when two diagonals are inclined at 30 and 5 respect-
ively ; (l>) when two alternate sides are inclined at 30 and
50 respectively ; (r) when AE and CF are inclined at 30 and
50 respectively.
5. Draw the plan of a regular tetrahedron, 2" edge, when a face
and an edge are inclined respectively at 55 and 40. Show
an elevation of the solid on a vertical plane which makes an
angle of 30 with the edge which is inclined at 40 .
6. Obtain the projections of an equilateral triangle, 2" side, when
two sides are inclined at 30 and 50 to the horizontal plane,
and the third side at 40 to the vertical plane.
7. Two lines meet at an angle of 60. The plane containing them
is inclined at 40 , and one of the lines is inclined at 50. Find
the inclination of the other. '
298 PRACTICAL SOLID GEOMETRY chap.
255. Problem. To draw the plan of a square, having
given the inclinations a t and a., of two of its sides.
Let AB, AD be the sides of the square which are in-
clined respectively at a x and a. 2 , and suppose the point A
to be on the ground.
Commence the solution by drawing a square A B Q C D ;
let this be the rabatment of ABCD from the required
position, into the ground, about the horizontal trace of the
plane containing the square.
The horizontal trace must now be found ; it is a line
through A such that if the rabatted square be turned back
about this trace, until AB be at the inclination a v then at
the same time AD shall be at the inclination a 2 .
Draw the right-angled triangles A 1 B x b v A x D x d v having
the base angles respectively equal to a x and a. and the
sides A X B V A X D X each equal to the side of the square ;
draw D X H X parallel to the base. Then, as in Art. 179,
B x b x , D x d x are the heights of B and D above A, i.e. above
the ground ; and A x b v A x d x are the lengths of the plans of
AB, AD. Also H x in A l B 1 determines the position of a
point Zfin AB, 'which is at the same height as D.
Make A Q If Q = A X H V and join D Q H . Then D If is
the rabatment of a horizontal line in the plane of the square.
The horizontal trace of the plane of rabatment is therefore
a line through A parallel to D If ; it is denoted by th in
the figure.
Finally, the required plan may be obtained in two ways.
First Method. With centre A () , radii A x b x , A x d x , the
lengths of the plans of AB, AD, describe arcs intersecting
the lines dawn through B , D Q , perpendicular to th, in b
and d respectively ; join ab, ad, and draw dc, be parallel to
ab, ad. Then abed is the required plan of the square.
B b, D d are the plans of the circular paths of B and D
traced during the rabatment.
Second Method. Take an xy perpendicular to D Q II Q , or
/// ; draw the two horizontal lines mm, nn, at heights above
xy equal to B x b v D x d^ respectively. With centre /, describe
XII
FIGURES IN GIVEN POSITIONS
299
arcs through B^,D^ (projected from fi , D { ), intersecting the
horizontal lines mm, nn in // and d' respectively. Then //, d'
are the elevations of B, D, a' is the elevation of A, and a,
d', b' will lie in one straight line, which is the edge eleva-
tion of the plane of the square. The arc C Q 'c determines
c ; and the plan of the square is obtained by drawing pro-
jectors from the points in elevation, to intersect lines from
the corresponding points of the rabatment, drawn at right
angles to th, the horizontal axis of rotation.
A model may be made to illustrate this problem.
Draw a square ABCD, Fig. ((7), and draw the lines Ab, Ad, making
angles of a v a 9 with AB, AD. Draw Bb, Dd perpendicular to Ab,
Ad. Make bo equal to Dd, and draw oH parallel to bA. Cut out
this figure in paper, then indent and fold the triangles ABb, ADd
about AB, AD to such a position that when Ab, Ad rest on the ground,
b and d are the plans of B and D.
By studying this model, the reasons for drawing the
various construction lines should be quite evident.
300 PRACTICAL SOLID GEOMETRY chap.
256. Problem. To draw the plan of a regular hexagon
of given side : (a) when the inclinations of two diagonals
are given ; (b) when the inclinations of two alternate sides
are given.
Draw A B Q C D E (j F , the rabatment of the hexagon.
(a) The rabatment C Ff Q of a horizontal line is found
with the aid of Fig. (a), as in Prob. 255. The horizontal
trace of the plane of rabatment is parallel to F Q H , and
may be taken through Q .
The plan is then determined as in Prob. 255, either
by the first or second method. If the first method be
adopted, and the plan of OAF be thus found, the plan of
the hexagon can be completed, without drawing an eleva-
tion, by making use of the following proposition of pure
solid geometry : "If a number of straight lines are parallel
to each other and of equal lengths, then their projections
on any plane are also parallel to each other and of equal
lengths P
Further, the elevation on any xy can be obtained with-
out first drawing the edge elevation, by making use of the
properties of the perpendiculars A x a v B x b v as explained in
Art. 179.
(b) Let the sides AF, DE be those having the given
inclinations. Produce them to meet in P.
Then by the aid of Fig. {I?) the rabatment F> A' of a
horizontal line in the plane of the hexagon is determined.
The horizontal trace of the plane of rabatment is parallel
to DqKq, and may be taken through F Q .
The plan can be obtained by either of the methods of
Prob. 255, making use of the proposition stated above if
the first method be the one adopted.
257 Problem. An equilateral triangle has two of its
sides AB, AC inclined at a x and a 2 to the horizontal plane,
and the third side BC inclined at 6 to the vertical plane ;
determine its plan and elevation.
Obtain the plan as in Prob. 255, by first drawing the
xir
FIGURES IN GIVEN POSITIONS
301
rabatment A B Q C ; then by the aid of the triangles A l B l b v
^\C\ C \ f m d C H the rabatment of a horizontal line; and
finally determine the plan abc by the first method.
To find the elevation, first draw a right-angled triangle
B 2 C/,, where B 2 C. 2 = BC, and the angle C 2 B 2 c 2 = ft. With
centre c, radius C\/ 2 , the difference in the distances of B and
C from the vertical plane (Art. 179), describe the circle
shown, and draw the tangent bm.
Then the plan abc and an elevation on an xy parallel to
bm will satisfy the conditions. Compare with Prob. 1S0.
302 PRACTICAL SOLID GEOMETRY chap.
258. Problem. Two lines meet at an angle of 65 '. The
plane containing them is inclined at 50 , and one of the
lines is inclined at 40 ; find the inclination of the other.
Represent a plane inclined at 50, in which place a line
AB inclined at 40 , and find its rabatment A B .
Draw a line A C making 65 with A B .
Determine the plan and elevation of AC when the plane
of rabatment has returned to its original position.
Then find the true inclination of AC.
259. Problem. Draw the plan of a regular tetrahedron,
\V edge, three of its corners heing at heights of -]-", \\",
and 11" respectively ahove the ground.
Draw A Q B Q C Q d x the plan of the tetrahedron, with the
face ABC on the ground, and consider this the plan of the
solid in its rabatted position.
In the two triangles shown to the right, make A 1 B 1 =
A l C 1 = 1 1", the length of edge of the solid, and B x b v C x c x
respectively equal to 1-}" and 1", the heights of B and C
above A. Draw C X H X parallel to c x A v and make A Jd =
A X H X ; then C Q J7 is the rabatment of a horizontal line in
the face ABC.
Take xy perpendicular to C Q Jif Q , and determine a'b'c the
elevation of ABC, as in Prob. 255, second method, assum-
ing the point A in the ground, and drawing the horizon-
tal lines nun, /in, at distances equal to B x b x , C x c x , above xy.
To determine the elevation of D, first find d x B> () , the
distance of D from the face ABC, by the rabatment of
CD about its plan as in the figure, or by any other method.
Obtain the point s as shown, and draw s'd' perpendicular
to db' and equal to d x B> Q . Then d' is the elevation of D,
and the elevation of the tetrahedron is completed by draw-
ing the lines representing the edges.
The plan is to be obtained as in previous problems, and
is left as an exercise for the student.
Finally, if an xy be drawn parallel to the one shown, and
|" below it, all the conditions of the problem are satisfied.
XI f
FIGURES IN GIVEN rOSITl<>\-
.-301
H./-\~tC,
b. c>
Examples. 1. Determine the plan of a hexagon of f " side, when
three alternate angular points are i", i|", and if" high re-
spectively.
2. Draw the plan of a square 2^" side, when the heights of its
centre and two corners, not opposite each other, are i", if",
and 2V respectively.
3. Draw the plan of an isosceles triangle, base 2^", sides 2", when
the extremities of the base and the middle point of one of the
sides are at heights of f", 1^", and 2" respectively.
4. Draw the plan of an equilateral triangle, 3" side, when the
three middle points of the sides are at heights of 1", l%", if"
respectively.
5. Draw the plan of a regular tetrahedron, 2\" edge, the heights of
three of its corners above the horizontal plane being respectively
2 ' *~2 ' ^
6. Draw the plan of a cube, 2^" edge, when three of its angular
points are at heights of 1", 1.25", and 2" above the horizontal
plane. Make an elevation on a plane parallel to one of the
diagonals of the solid.
7. An isosceles triangle is the plan of an equilateral triangle. Find
the inclination of the plane of the triangle (1) when the equal
sides are each three-fourths of the base ; (2) when the base is
three-fourths of each of the equal sides.
8. Draw the plan of an octahedron of 2" edge, when two diagonals
of the solid are inclined at 26 and 36 respectively.
9. Three corners of a square, 2" side, are 1", 1.4", and 2.1" high ;
find the height of the centre.
304 PRACTICAL SOLID GEOMETRY chap.
260. Problem. To draw the plan of a cube having
given the length of edge and the inclinations e and 9 of
two faces.
Draw the traces vth of a plane inclined at 0.
By Prob. 233 determine a plane at right angles to VTH
and inclined at <f>. That is, choose any point A in the .
plane and draw the projections of a cone, vertex A, base
on ground, base angle <. Find also the horizontal trace
JV of a line through A perpendicular to the plane VTH
Then the tangent nlm is the horizontal trace of the
plane at <.
The line ias is the plan of the intersection of the planes.
If now one edge of the cube coincide with I A, and two
faces with the two planes, the conditions will be satisfied.
Rabat the plane VTH, and thus obtain t'S .
Draw the square with one side in t'S ; this is the plan
of the cube while on the ground. Now let the plane VTH
be raised to its proper position, and draw the corresponding
plan and elevation of the cube.
261. Problem. To draw the plan of a tetrahedron,
having given the length of edge and the inclinations e
and 9 of two faces.
Draw the traces vth of a plane inclined at 6.
Then by Prob. 234 determine the traces of a plane
which is inclined at <, and makes with the plane VTH an
angle a equal to that between the base and one of the equal
faces of the tetrahedron.
Determine the intersection of these two planes and
rabat it, along with the inclined plane, into the ground.
Proceed now exactly as in the last problem, by drawing
an equilateral triangle having one side (equal to the given
edge) in the rabatment of the intersection.
While the inclined plane is in the ground complete the
plan of the tetrahedron. Finally suppose the inclined
plane to be raised to its proper position, and thus complete
the required plan and elevation of the tetrahedron.
xn
FIGURES IN GIVEN POSITIONS
3o5
X-A
\\ /--' V.
/,
306 PRACTICAL SOLID GEOMETRY chai\
262. Problem. To draw the plan of a cube, having
given the length of edge, the inclination e of one face, and
the inclination 9 of a diagonal of the solid. (No figure.)
There are three preliminary problems to be worked.
1. Find the length of the diagonal.
2. Find the angle a between a diagonal and a face.
3. Find the angle /3 between a diagonal and an edge.
Having solved these, draw the projections of a diagonal
inclined at <f>. Since a plan only is required, the diagonal
may be parallel to the vertical plane.
Then by Prob. 235 determine a plane inclined at 6, and
making an angle a with the diagonal.
Next obtain the plan of an edge which meets the dia-
gonal and lies in the plane. This is done by Prob. 236, the
edge making j3 with the diagonal.
Now rabat the plane with the edge into the ground.
On the rabatment of the edge construct a square. This
will be the plan of the cube while one of its faces is on the
ground.
Now turn the plane back into its original position, carry-
ing the cube with it, and thus complete the plan of the
solid by well-known methods.
Examples. 1. Draw the plan of a cube, 2^" edge, when two of
its adjacent faces are inclined at 45 and 75 respectively.
2. A building brick 9" x 4^" x 3" has one end inclined at 40, and
a long face at 6o. Draw its plan. Scale \.
3. A tetrahedron, 2^" edge, has two of its faces inclined at 40
and 70. Draw the plan and elevation.
4. A square pyramid, base 2" side, axis 3" long, has its base in-
clined 40 , and one of its faces at 6o. Draw its plan, and a
sectional elevation on a vertical plane which bisects any two of
its long edges.
5. A cube, 2\" edge, has a diagonal inclined at 6o, and a face at
65 ; determine its plan.
6. Draw the plan of a regular tetrahedron of 2" edge, when the
line joining a vertex to the centre of the opposite face is hori-
zontal, one of the other faces being inclined at 6o. What is
the least angle which may be substituted for the 6o, so as not
to make the solution impossible ? Ans. \0)\.
xii FIGURES IN GIVEN POSITIONS 307
263. Miscellaneous Examples.
1. Draw the plan of a hexagon of 1.5" side, the heights of three
successive adjacent corners to be 1", 1.5", and 0.75". (1878)
2. An isosceles triangle (sides z\" , base i|") has its base in the
horizontal plane and one side in the vertical plane. The base
makes an angle of 35 with xy. Determine the plan and eleva-
tion of the triangle. ( J 893)
3. A square of 3" side lies on a plane inclined at 50 . One side of
the square makes 40 with the horizontal trace of the plane.
Draw its plan. (1885)
4. Draw the plan of a cube of i T V' edge, one face inclined at 50
and a second face at 60 . (1890)
5. Draw the plan of an octahedron of 2" edge, one edge being in-
clined at 30 , and another (on the same face) at 20 . (18S2)
6. Draw the complete plan of a cube of 1.5" edge, two faces in-
clined respectively at Co Q and 70 to the horizontal plane.
(iS95)
7. A right pyramid has for its base a regular pentagon of which the
diagonals measure 2.5". The vertex is 2" above the base.
Draw the plan and elevation of the pyramid, with its base in a
plane inclined at 55 to the vertical plane, and at 6o to the
horizontal plane ; one diagonal inclined at 30 , and one end of
that diagonal in the vertical plane. (1894)
8. Draw a plane inclined at 45 to the horizontal plane, and at 6o
to the vertical plane. Draw a line in the plane inclined at 30 ,
and 2" long between its traces. Lastly, draw the plan of a
regular hexagon lying in the plane of which the above line is a
diagonal. O897)
9. An octahedron of 2j" edge has the plane containing two of its
diagonals inclined at 30, and that containing one of these, and
the other diagonal inclined at 70 . Draw its plan.
10. Determine the projections of a cube on three planes mutually
perpendicular, having given the inclinations of three adjacent
edges, one to each plane.
11. Determine the projections of an equilateral triangle on three
planes mutually perpendicular, having given the inclinations of
its sides, one to each plane.
12. Draw the projections of a cube on three planes mutually per-
pendicular, having given the inclinations of three adjacent faces,
one to each plane.
CHAPTER XIII
THE PROJECTION OF CURVES AND CURVED SURFACES
264. General method of procedure.- In the problems
hitherto considered, the figures represented in projection
have been made up of points, straight lines, and planes.
The projection of curves, and of solids bounded by curved
surfaces, is now to be considered.
The general method of projecting a curve is to first
find the projections of a number of isolated points in it ;
then to draw a curve by freehand through the points thus
determined.
In projecting a solid, the projections of the straight or
curved edges are first found ; then if necessary the outline
of the projection is completed, as determined by projectors
which touch the surface of the solid. This will be more
particularly explained in the problems which follow.
265. Problem. Draw the plan of a circle, \\" diameter,
when its plane is inclined at 50\
First Method. Begin with the rabatment, a circle
A B , i\" diameter. Take twelve equidistant points on
the circumference of this circle, as shown. Find the plans
and elevations of these points when the plane of rabatment
is turned back into its original position 7'th, inclined at 50.
The line ab' is the edge elevation of the circle ; and a
fair curve carefully drawn through the plans of the points is
the plan of the circle. The plan is an ellipse.
CH. XIII
CURVES AND CURVED SURFACES
309
o
Second Method. Begin by drawing an edge eleva-
tion of the circle, the line a'b', 1^" long, inclined at 50'.
On a'b' draw a semicircle, which divide into six equal
arcs as shown. From the points of division draw lines
perpendicular to a'b', intersecting the latter in o', 1', 2 '.
Conceive o', 1', 2 as the elevations of chords of the
circle perpendicular to the vertical plane. Then the dotted
perpendiculars represent the halves of these chords, when
half the circle is turned about AB into a position parallel to
the vertical plane.
To obtain the plan of the circle, draw projectors from
the points in a'b'. Consider the projector from 1 ' : let m
be its intersection with ab ; make mi, m\ each equal to
i'ij^; repeat this construction for the other points. In this
way the points o, 1, 2 on the plan of the circle are found,
and the ellipse is drawn through them.
310 PRACTICAL SOLID GEOMETRY chap.
266. Problem. A given sphere resting on the ground
is cut by a given vertical plane. Draw the plan and
elevation of the trace of the section plane on the surface,
and show a sectional elevation of the sphere.
The projections of the sphere are circles, equal in
diameter to the sphere, drawn with centres c, c, the plan
and elevation of the centre of the sphere. Since the
sphere rests on the ground, the elevation touches xy.
Let VTH be the given vertical section plane. The
section is a circle of which ab is the plan. The elevation
may be found as follows :
On ab as diameter draw a semicircle, and divide its
circumference into six equal arcs as shown. From the
points of division draw lines perpendicular to ab, intersect-
ing the latter in o, i, 2.
Let o, 1, 2 be the plans of vertical chords of the circle.
Then the dotted perpendiculars represent the halves of
these chords, when half the circle is turned about AB into
a horizontal position.
Draw projectors from the points on ab. Consider the
projector from 2 : let ;;/ be its intersection with the hori-
zontal line through c ; make 1112, 1112 each equal to 2 Q 2 ;
repeat this construction for the other points, and through
the points in elevation thus found draw a curve, which is
an ellipse, and is the elevation of the required trace of
the section plane on the surface of the sphere.
The sectional elevation is that on xy\ taken parallel to
ab. The method of obtaining this needs no further ex-
planation.
Examples. 1. Draw the plan of a circle 2^" diameter, when its
plane is inclined at 50 .
2. Draw the elevation of a circle 2\" diameter whose plane is
vertical and inclined at 60 to the vertical plane.
3. A hemisphere, 2^" diameter, rests with its flat face on the ground.
It is cut by a vertical plane h" distant from its centre, which
makes an angle of 50 with xy. Draw the elevation, showing
the curve of intersection.
XIII
CURVES AND CURVED SURFACES
3ii
6.
A sphere 3" diameter is cut into four equal parts by two
perpendicular planes through its centre. Draw the plan of
one of these parts when resting with a flat face on the ground ;
and give a sectional elevation on a vertical plane which makes
45' with the straight edge of the solid, and is distant 1"
from the centre.
A sphere, 4" diameter, is cut into eight equal parts by three planes
mutually perpendicular. Draw the plan of one of these parts
when resting on a flat face. Also draw an elevation on a
vertical plane, which makes an angle of 30 with a horizontal
edge of the solid, so as to show the flat faces.
Draw the plan of the solid of Ex. 5 when two of its edges are
inclined at 30 and 4o\ And add an elevation on a vertical
plane, the real angle between which and the third edge is 25.
Draw the plan of the solid of Ex. 5 when its curved surface rests
on the ground, two faces being inclined each at 70 .
312 PRACTICAL SOLID GEOMETRY chap.
267. Problem. A cone, diameter of base 1\", length of
axis 2 ", rests with its base on the ground, and is cut by a
plane inclined at 45, bisecting the axis, (a) Draw the
plan and elevation of the cone, showing the trace of the
section plane on its surface ; (b) find the true shape of the
section ; (c) obtain a development of the surface of the
cone, showing the trace of the cutting plane on the surface.
(a) The plan of the cone is the circle ab, with s, the
plan of the veitex, as centre. The elevation is the triangle
sab.
To find the section, take twelve equidistant points on
the circumference of the base, join these to the vertex, and
draw the plan and elevation of the twelve generating lines
thus formed. The intersection of the cutting plane with
these generators gives twelve points on the curve of section
required. The elevations of the points are in c'd', and the
plans are found by projecting from the points in elevation.
A special construction is required to determine the plans
o, o ; in this case it is evident that so, so are each equal to
o'o", where do" is parallel to xy. The curve through the
plans of the points is an ellipse.
(b) The true shape of the section may be found by a
rabatment of the cutting plane about its vertical trace, which
is equivalent to an auxiliary plan on tv. The rabatment
of CD is c x d v and to obtain the points on the curve, make
m t i v in, i, each equal to i/n ; and repeat this construction
for the other points. The true shape of the section is an
ellipse.
(c) To obtain the development. With S as centre,
describe an arc with radius equal to s'd, the length of a
generator.
Take any point A on this arc, and by Rankine's con-
struction, Prob. 113, make arc AT equal to \ circumference
of base of cone; and set off TB = AT. Subdivide each
of these arcs into three equal parts. The development of
the curved surface of the cone, with that of the twelve
equidistant generators, is thus obtained.
XIII
CURVES AND CURVED SURFACES
3i;
vvy
Make SO = so" = true distance of the vertex from the
point O ; and repeat this construction for the other points.
A curve through these points is the development of the
trace of the cutting plane on the surface of the cone.
Examples. 1. Determine the true shape of the section of a cone
2h" base, 3" axis, by a plane parallel to its axis, the distance
between the plane and the axis being j". Develop th cone
and show the curve of section.
2. Determine the shape of the section of a cone 2" base, 3" axis, by
a plane parallel to a tangent plane, ap.d distant J" from the
latter.
3. A cone of which the altitude = radius of base= ii" is cut in two
by a plane containing the vertex, the trace of the plane on the
base bisecting a radius at right angles. Draw the plan of the
smaller part when resting with its triangular face on the ground ;
and obtain an elevation on a ground line which makes an angle
of 40 3 with the chord of the segmental base.
314 PRACTICAL SOLID GEOMETRY chap.
268. Problem. A cone, diameter of base \\", length of
axis 2", rests with a generator on the ground, the axis
being parallel to the vertical plane. Draw the plan and
elevation of the cone ; and add a sectional elevation on a
vertical plane which bisects the plan of the axis at an
angle of 45.
The elevation of the cone is the triangle s'db\ copied
from the triangle sab' of Fig. 278, but with the generator
SA on the ground.
The plan of the circular base is an ellipse determined as
in Prob. 265, second method; and the plan of the cone
is completed by drawing the two tangents to the ellipse
from s.
Each of these tangents is the line generated by the foot
of a vertical projector, which projector moves so as always
to touch the curved surface of the cone. It may be shown
by pure solid geometry that the moving projector touches
the surface along two generators, one on each side of the
cone ; the two tangents from s to the ellipse are the plans
of these generators.
Let ////, bisecting sc at 45 , be the plan of the vertical
section plane. To obtain the sectional elevation, take a
new ground line x'y parallel to Im, and find the elevation
on x'y of the vertex and the points on the base according
to the rules of Art. 170.
Draw the twelve generating lines in both plan and
sectional elevation, whence twelve points on the section are
at once found by projection from the plan.
The drawing of this view is left as an exercise for the
student.
Examples. 1. A cone, 2" base, 3" axis, rests with a generator on
the ground ; draw its plan, and a sectional elevation on a vertical
plane which bisects the plan of the axis at 45 .
2. Draw the plan and elevation of a cone 2" base, 3" axis, when
the axis is inclined at 30 to the horizontal plane and 45 to
the vertical plane. Show the section by a horizontal plane
which bisects the axis.
XIII
CURVES AND CURVED SUREACES
315
Examples on Problems 270 to 272.
1. A cylinder, i}/ diameter, 2" long, has a cone 2f" base, i^"
axis, placed centrally on one of its ends. Draw the plan of
the solid when the bases of both cone and cylinder touch
the ground. Draw a sectional plan on a plane which con-
tains the vertex and two points on the circumference of the
base of the cone 2^" apart.
2. Draw a semicircle 5" diameter, in which inscribe the largest
possible circle. The semicircle is the development of a cone,
and the circle that of a line on its surface. Draw the plan and
an elevation of the cone when resting with its base on the
ground, showing the line on its surface.
3. Draw a sector of a circle 3" radius, containing an angle of 120 .
This sector is the development of the surface of a cone. Deter-
mine the plan and elevation of the cone when resting with its
base on the ground.
Suppose a fly, starting from a point in the circumference of
the base, to walk round the surface and return to the same
point. Show the plan and elevation of the path, when its
length is the least possible.
316 PRACTICAL SOLID GEOMETRY chap.
269. Problem. A cylinder, 2" diameter, is 3" long.
Draw the plan and elevation : (a) when the axis is parallel
to the vertical plane, and inclined at 45 to the horizontal
plane ; (b) when the axis is inclined to both planes of pro-
jection.
(a) Draw the rectangle a'a'b'b' having b'b'=$'\ b'a =2",
and b'b' making an angle of 45" with xy. Draw c'c bisecting
the short sides of the rectangle. Then a'a'b'b' is the eleva-
tion of the cylinder, and c'c that of the axis ; a'b\ a'b' are
the elevations of the two circular ends.
The plan is obtained by projecting the two circular ends
from the elevation, in the manner of Prob. 265, second
method. The two ellipses thus found are joined by two
common tangents as shown in the figure, and the plan
is complete.
(b) An elevation on any line xy\ not parallel to cc,
determined according to the principles of Art. 170, will
satisfy the requirements of (b). The two ends project into
ellipses, common tangents to which complete the elevation.
The common tangents to the ellipses, and the lines
a'a', b'b', are the projections of the generators along which
the projectors touch the curved surface of the cylinder in
each case.
Examples. 1. A cylinder, 2" diameter, is 2|" long, draw the
plan and elevation : (a) when the axis is parallel to the vertical
plane, and inclined 40 to the horizontal plane ; (b) when the
axis is inclined at 50 and 40 to the vertical and horizontal
planes of projection.
2. A cylinder, 2V base, 3^" long, is cut into two parts by a plane
containing the axis. Draw the plan of one of these parts, when
resting with a rectangular face on the ground, and draw a
sectional elevation on a vertical plane bisecting the axis at an
angle of 45 .
3. A cylindrical cheese is 16" diameter, and 5" thick. A wedge-
shaped piece is cut out by two planes containing the axis and
including an angle of 50 . Draw the plan of the piece when
resting with one of its rectangular faces on the horizontal plane,
and draw an elevation on a plane making 25 with the short
sides of that face. Scale ^.
XIII
CURVES AND CURVED SURFACES
317
270. Problem. To find the projections of the shortest
line on a given developable surface, connecting two given
points on the surface.
First read Prob. 271. The line in question develops
into a straight line, joining the developments of the points.
For the length of the line on the surface is equal to the
length of its development, and when the development is
straight it is the shortest line possible between the points.
Therefore, first find a development of the given surface
and of the given points in it. Draw the straight line between
the developed points. Then working backwards from
the development, determine the projections of the line.
See Fig. 271.
Note. The shortest line between two points on a cylinder is a helix
of uniform pitch ; for the development of this curve is a straight line.
See Probs. 369 and 130.
318 PRACTICAL SOLID GEOMETRY chap.
271. Problem. The given sector of a circle SBAB is a
development of the curved surface of a cone ; and HK
that of a line on the surface. Draw the plan and an
elevation of the cone, when resting with its base on the
ground, showing the line HK.
The diameter of the base of the cone must first be
found. Bisect the arc BB in A, and the arc AB in T.
Draw the tangent AR and make AR equal to the chord of
\ the arc AT. With centre R describe an arc through T,
intersecting in / the line At drawn at 45 to SA produced.
Draw tC perpendicular to SC. Then the circle with centre
C, radius CA, is the development of the base of the cone.
For by Rankine's construction, Prob. 113, arc At arc AT
="| .circumference of base.
The plan of the cone is thus drawn ; and the altitude
and the elevation may be found, since the diameter of the
base, and the length of a generator, viz. SB, are known.
To obtain the plan of HK, take a series of generators
intersecting the line. In the figure, four of twelve equi-
distant generators are shown. The construction is then
equivalent to that of Prob. 267 (c), worked the reverse
way, and the completion is left as an exercise.
272. Problem. The plan and elevation of a solid are
shown in the figure at (a). Draw the plan when the solid
is in the position shown in elevation at (b).
The plan of the cylindrical portion of the solid is found
as in Prob. 269.
The curves on the four side faces of the solid are equal
and similarly placed circular arcs, being the intersections of
the faces with the spherical surface. The points A, B, C
on these arcs are similarly and symmetrically taken on the
four faces. Hence the distance between b, b in plan is the
same as between b', b' on the front face in elevation. This
consideration enables the plans of the arcs to be found.
The plan of the solid is completed by projecting from the
elevation the edges which are straight.
XIII
CURVES AND CURVED SURFACES
319
320 PRACTICAL SOLID GEOMETRY chap.
273. Some properties of the cone and cylinder. Before
proceeding with the remaining problems of this chapter we
wish to amplify some of the theorems which are given in
the Appendix.
Sphere inscribed in cone or cylinder. Take two inter-
secting lines of indefinite length and draw the bisector of
the angle between them with any point on this bisector as
centre draw the circle touching the lines. Suppose now
these lines and the circle to rotate about the bisector as
axis, then it is evident that the lines will generate a cone,
and the circle a sphere inscribed in the cone, and the sphere
will touch the cone in a circle perpendicular to the axis.
It is readily seen that an indefinite number of spheres of
varying sizes can be inscribed in a cone, all having their
centres in the axis.
A cylinder, being the limiting case of a cone where the
vertex is at an infinite distance away, may also have an
unlimited number of spheres inscribed in it. All these
spheres will have a diameter equal to the diameter of the
cylinder.
Two cones and a common inscribed sphere. It can be
shown that if two cones circumscribe the same sphere, their
surfaces intersect each other in two ellipses. Also, since a
cylinder is a particular case of a cone, a similar remark
applies to two cylinders, and to a cylinder and cone. This
property may be frequently taken advantage of to facilitate
the working in certain problems.
Projection of a cone or cylinder of indefinite length. A
cone of indefinite length may evidently be defined by any
two spheres inscribed in it, for if these two spheres are
given, then the line of the axis, the position of the vertex,
and angle of the cone are known. One of the spheres may
be at the vertex and so become a point ; thus a cone of
indefinite length is defined by its vertex and an inscribed
sphere. The projection of the cone on any plane consists
of the two tangents drawn to the two circles which are the
projections of the two inscribed spheres. It must be
XIII
CURVES AND CURVED SUREACES
321
specified whether the two external or two internal tangents
are to be taken. As before, one of these circles may be a
point, i.e. be the vertex of the cone. Similar remarks apply
to a cylinder of indefinite length, but now the two inscribed
spheres are of equal size. A cylinder and cone projected
on the horizontal and vertical planes are illustrated in
the figure. These indefinite cones and cylinders may be
drawn broken at the ends as shown.
These statements must be qualified as follows. If the
axis of the cylinder be perpendicular to the plane of
projection, the projection of the cylinder is a circle which
is the edge view of the surface.
If the projection of the vertex of the cone fall within the
projection of an inscribed sphere, the common tangents
cannot be drawn ; in this case the indefinite cone has no
definite form of projection, but it may still be represented
in projection by the circle and point or by two circles.
Y
322 PRACTICAL SOLID GEOMETRY chap.
274. Problem. To determine the horizontal trace of a
given cone of indefinite length, whose axis EV is parallel
to the vertical plane.
The cone may be given by its vertex V, vertical angle a,
and inclination of axis 6 ; or by its vertex and an inscribed
sphere. In either case the projections in outline are readily
drawn ; let them be as shown in the figure. The horizontal
trace is a conic section, in the present case an ellipse.
To find the major axis of the ellipse. Let the outline in
elevation intersect xy in a', a^ ; bisect a'a^ in c . Project
a', a x ' and c on to the plan of the axis at #, a v and c. The
horizontal trace of the cone consists of an ellipse of which C
in the centre, and AA 1 the major axis.
To set out the ellipse. First Method. Bisect the angle
v'a'a^ by the line a'i' ; draw i'f perpendicular to xy, and
with i' as centre and i'f as radius draw the circle shown ;
this circle is the elevation of a focal sphere of the cone with
respect to the ground as section plane (see Art. 76). There-
fore the point of contact/' is the elevation of one focus of
the ellipse, and fi, projected from/', is its plan. Set off a.f x
equal to af v then f x is the second focus. The minor axis
of the ellipse may now be found and the ellipse drawn,
as explained in Chapter IV.
Second Method. Through c, the middle point of a'a^,
draw h'k' perpendicular to v'e . On h'k' describe a semi-
circle, and draw e'b ' perpendicular to h'k'. Then c'b ' is the
length of the semi-minor axis, and the horizontal trace of
the cone can now be completed as before.
To explain this construction, observe that c', being the
mid-point of a x 'a, must be the elevation of the minor
axis, the minor axis itself being a chord of that circular
section of the cone which has h'k' for its elevation. One-
half of this circle is turned about HK until it is parallel
to the vertical plane ; its elevation then being the semi-
circle on lik' ; the semi-minor axis appears as c'b '.
Third Method. Repeat the construction of the second
method for a number of points in a'a x ' in addition to the
xii r
CURVES AND CURVED SURFACES
32:
mid-point c . The lines corresponding to c'b^ thus obtained
will be ordinates of the elliptic trace, and may be set off on
each side of va v on the projectors from the respective
points in a'a(. This method is useful when the cone is so
situated that its trace is a parabola or hyperbola; or an
ellipse so elongated that the end A x is inaccessible.
Fourth Method. Draw the plan of the focal sphere or
any other sphere inscribed in the cone, and from v draw
tangents to this circle. These tangents give the plan of the
cone in outline and must touch the trace (see Art. 284).
The trace can therefore be drawn, since it is an ellipse, of
which the major axis and a tangent are known (see Prob. 98).
324 PRACTICAL SOLID GEOMETRY chap.
275. Problem. To determine the horizontal trace
of a given cylinder of indefinite length, whose axis ED
is parallel to the vertical plane.
The cylinder may be given by its diameter, the position
of a point D in its axis, and the inclination 6 of the axis ;
or by two inscribed spheres as explained in Art. 273. In
either case the plan and elevation of the outline of the
cylinder are readily determined. Let them be as shown in
the figure.
The horizontal trace of the cylinder, being a plane
section, must be an ellipse. To determine the plan of the
ellipse we may proceed as follows :
Let the outline in elevation meet xy in a'a t ' ; bisect
a'aJ in /. Project a, a' v c on to the plan of the axis at
a, a v c, and let the projector from c intersect the plan of
the outline in b, b y Then c is the centre, and aa v bb : the
major and minor axes of the elliptic trace. The ellipse can
be drawn by any of the methods described in Chapter IV.
The elevation of the trace is aa^.
The foci of the ellipse are shown in the figure ; these
might have been determined by a focal sphere, as was
explained for the cone in the last problem.
Examples. 1. A circle |" diameter, centre s, is the plan of a sphere
which rests on the ground. A point v, 1" from s, is the plan
of the vertex of a cone of indefinite length which circumscribes
the sphere, the height of /"being iV'. Draw the plan of the
cone and an elevation on a vertical plane parallel to VS. Set
out the horizontal trace of the cone.
2. A cylinder, 2" diameter, has its axis parallel to the vertical
plane and inclined 50 to the ground. Determine its horizontal
trace.
Show the elevation of the section made by a vertical plane
parallel to, and distant i-" from the axis.
3. Draw a triangle a! b'c with b'c' in xy ; make b'c' 3", a'b' 4",
and a'c 2^". This is the elevation of a cone, the plan of whose
axis makes 40 with xy. Determine the horizontal trace of
the cone. See Prob. 276.
4. A cylinder 2" diameter has its axis inclined at 45 to the ground
and 30 to the vertical plane. Determine its horizontal trace ;
also its vertical trace. See Prob. 277.
XIII
CURVES AND CURVED SURFACES
525
x a.
b
c
\CL
f !
a
k
5. Determine the trace of the cone of Ex. 2, p. 330, on a horizontal
plane J" above the ground.
6. Determine the section of the cone of Ex. 3, p. 330, by a hori-
zontal plane which touches the smaller sphere at its highest
point.
7. A cone, vertical angle 6o, has its axis horizontal and 2" above
the ground. Determine the section by a vertical plane (a)
which makes 45 with the axis and is distant 2" from the
vertex ; (/>) which makes 20 with the axis, and is distant A"
from the vertex.
8. Determine the vertical trace of a cylinder, ii" diameter, whose
axis makes 40 with each plane of projection.
326 PRACTICAL SOLID GEOMETRY chap.
276. Problem. Having given mVn', the elevation of a
cone, and also the projections of the axis EV, to determine
the horizontal trace of the cone.
Let it be observed that m'ri is not the true length of
the major axis, nor is it the length of the elevation of the
major axis of the required elliptical trace, for the axis of the
cone is not parallel to the vertical plane as in Prob. 274.
None of the methods there explained can be immediately used.
First Method. Obtain an auxiliary elevation of the cone
as seen when looking in the direction shown by the arrow
in plan, that is on x'y' taken parallel to ev. In determining
this elevation in outline use may be made of an inscribed
sphere, as explained in Art. 273. Then any of the methods
in Prob. 274 may be employed.
Second Method. Bisect the angle vi n'v by the line n'i'
and draw i'f perpendicular to xy ; then / is the centre of
one focal sphere, and F is the corresponding focus. Bisect
m'ri in c, and by projection from c and /' obtain c and_/ on
ve, produced if necessary. The point C is the centre of the
ellipse.
Draw the projector n'r\ then it is evident that n'r is a
tangent to the required ellipse.
Draw fh perpendicular to n'r, and with c as centre and
ch as radius describe the arc ha to meet ev in a ; then a is
one extremity of the major axis of the required ellipse, and
c? 1 is the other, where ca x is made equal to ca. The minor
axis can now be found as in Prob. 79, and the ellipse con-
structed. The plan of the cone is completed by drawing
tangents from v to the ellipse.
277. Problem. Having given the projections of cylinder
of indefinite length, to determine the horizontal trace of the
cylinder. (No figure.)
As in the last problem, the axis of the solid is to be
taken inclined to both planes of projection, and the outline
in plan and elevation may be drawn as tangents to the pro-
jections of two inscribed spheres.
XIII
CURVES AND CURVED SURFACES
327
The construction to find the trace is exactly like that for
the cone ; but in the case of the cylinder may be somewhat
simplified, because the minor axis of the ellipse is known,
being equal to the diameter of the cylinder.
328 PRACTICAL SOLID GEOMETRY chap.
278. Problem. The plan of a cone of indefinite length
is given, the plan of the axis being indexed ; determine
the index of p, the given plan of a point on the upper
portion of the cone.
First Method. Let mvn be the plan of the cone, and ev,
with the indices attached, the plan of the axis.
Draw xy parallel to ev, and project /, v . Describe a
circle to pass through p and touch vm, vn (Prob. 61);
let its centre be s. First regard this circle as the plan of
an inscribed sphere and determine its elevation, that is, the
circle with s' as centre. Complete the elevation of the
cone in outline by drawing the tangents v'k', v'k' .
Now regard the circle, centre s, as the plan of a vertical
cylinder, the elevation of which is indicated. Thus, the
cone and cylinder circumscribe the same sphere, and hence
(see Art. 273) their curve of intersection consists of two
ellipses. But it is evident, from considerations of symmetry,
that we shall obtain an edge view of these ellipses when
looking horizontally in a direction at right angles to both
axes ; and since /', ze/, z, u' are clearly the elevations of
points which are on both surfaces, it follows that t'w and
u'z are the elevations of these ellipses.
Now P is on the surface of the cone, and it has been
arranged that it is also on the cylinder, therefore it is on one
or the other of the two ellipses ; but from the condition that
P is on the upper half of the cone, it will be seen that P
is on the ellipse TW, and hence p' is on t'w'. The required
height of Pis therefore/'/, from which the index of P may
be measured.
See over leaf for the second method.
Note 1. It should be observed that /Ms not on the sphere, centre S,
although p is on the plan of this sphere. Draw k'k' joining the points
of contact of the tangents from v to the circle, centre / ; join p'v'
intersecting k'k' in /', then T is the point of contact of the generator
PV with the sphere, centre S. Hence if p'c' be drawn parallel to t's',
then C will be the centre of that inscribed sphere which contains P ;
c is the plan of the centre of this sphere. This will be required in
some future problems.
XIII
CURVES AND CURVED SURFACES
330 PRACTICAL SOLID GEOMETRY chap.
Second Method. Let mvn be the given plan of the
cone, egJ 1A being the indexed plan of its axis.
Draw any circle, centre s, to touch mv and nv ; this is
the plan of an inscribed sphere.
Draw vp, and regard it, not only as the plan of the
generator through P, but also as the plan of the vertical
plane containing this generator. Such a plane cuts the
sphere in a circle, diameter ao, to which the generator PV
is a tangent.
An elevation of this circle and tangent are shown on xy,
taken parallel to pv. In determining this elevation, the
heights of e and v are known from the given indices : s' is
the elevation of the centre of the sphere, and also of the
centre of the circle diameter ad.
Draw the tangent v'f. Then a projector from/ to meet
v't' produced in p\ will determine r'p' the height of P, and
the index of/ is thus known.
Note 2. It will be useful for the student to observe that the
generator VP touches the sphere, centre S, at T, and by obtaining t
from /', joining si and drawing pc parallel to st, we obtain the plan of
the centre of that inscribed sphere which contains /'. This sphere
will be required in some future problems.
Examples. 1. Describe a circle i-j" diameter, centre 5 ; take a
point v I J" from s ; these are the plans of an inscribed sphere
and vertex of a cone whose axis is inclined at 25 to the ground.
Take a point/ on the given circle, distant 2j" from v, and regard
this as the plan of a point on the surface of the cone. Determine
the height of /"above the ground, if that of the vertex be 2-\".
Determine the plan of that inscribed sphere of the cone
which passes through P.
2. A cone whose vertical angle is 90 , rests with a generator on the
ground ; draw its plan. Determine a point on the surface of
the cone which is ii" high and 2" distant from the vertex.
3. Two circles 2" and l|" diameter, which have their centres 1^"
apart, are the projections of two spheres, the heights of the
centres being respectively ii" and 1" above the ground.
Regard the points of intersection of the circles as the plans of
two points on the upper and lower surface of the cone which
circumscribes the spheres. Required the indices of the points ;
unit = 0.1". Also determine the inscribed spheres on the sur-
faces of which the points lie.
XIII
CURVES AND CURVED SURFACES
332 PRACTICAL SOLID GEOMETRY chap.
279. Problem. The plan of a cylinder of indefinite
length is given, the plan of the axis being indexed ; deter-
mine the index of p, the given plan of a point on the
lower half of the cylinder.
First Method. Let cd, with the indices attached,
be the plan of the axis of the cylinder. Take xy parallel
to cd, project cd', and draw the outline elevation of the
cylinder.
Describe a circle to pass through p and touch the out-
line of the given plan of the cylinder. First regard this
circle as the plan of a sphere inscribed in the given cylinder,
and then regard it as the plan of a vertical cylinder, the
elevation of which may be at once drawn. In this way it is
ensured that the two cylinders circumscribe the same sphere,
and therefore intersect each other in two ellipses. From
what was said in the last problem, it will readily be seen
that t'w' and u'z are the edge elevations of these ellipses,
and that p' will be on u'z , since P is on the lower half of
the cylinder. Measure p'r , and index / accordingly.
Note i. Since any number of points may be taken on the plan of
the vertical cylinder in either this problem or the last, and the corre-
sponding elevations determined, it follows that the projections of any
number of generators of the given cone or cylinder may be deter-
mined ; and by finding their horizontal traces the horizontal trace of
the cone or cylinder may be determined, as the fair curve through
these points. This constitutes a method of setting out the trace of a
cone or cylinder, additional to the methods given in Probs. 274
to 277.
Second Method. By regarding a cylinder as the
particular case of a cone when the latter has its vertical
angle diminished indefinitely, it will be seen that the second
method of the previous problem (p. 330) will apply to the
case of a cylinder, the generator through P being now
parallel to the axis.
Note 2. The sphere inscribed in the cylinder and containing P
may be determined as in Note 2 of the last problem.
XIII
CURVES AND CURVED SURFACES
333
s
\
Examples. 1. Draw a line ab i^" long ; a w b 22 is the indexed plan
of the axis of a cylinder 2" diameter. Select any point/ j"
from ab this is the, plan of a point on the surface of the
cylinder. Required the index of the point p. Unit = 0.1".
Also determine the indexed plan of the inscribed sphere which
passes through P.
2. Determine a sectional elevation of the cylinder of Ex. 1 on
a vertical plane whose horizontal trace makes 60 with ab.
3. Copy the above figure double size ; then draw the elevation of the
ellipses TIV, UZ on a vertical plane which makes 45" with xy.
334 PRACTICAL SOLID GEOMETRY chap.
280. Problem. The plan of a cylinder of indefinite
length is given, gd, the plan of the axis, being indexed.
Determine the true shape of the section made by the
vertical plane, the horizontal trace of which is lm.
Bisect the angle nqr by the line qi x ; draw /,/ at right
angles to lm, then I x is the centre of a focal sphere with
regard to the section plane LM, and F x is the corresponding
focus. From symmetry F is the other focus where rf= qf v
Draw xy parallel to lm and obtain g ', d' by projection
from g, d, setting off the heights given by the indices.
Now since LM is a vertical plane, the radius L l F l of the
sphere is horizontal ; and I x is on the axis of the cylinder.
Hence z' a ' and /' coincide, and each is on g'd' .
Project/' and/' from/ and/ Bisect//' at right angles
by the line b'by, and make c'b' = c'b{ = radius of cylinder;
then b'b^ is the minor axis. The major axis a'a^ may now
be determined and the ellipse may be completed by any of
the methods explained in Chapter IV.
281. Projection of any surface of revolution. Reasoning
as in Art. 273, it is readily seen that a series of spheres can be
inscribed in any surface of revolution. The surface itself
may be conceived as generated by, or the envelope of, its
inscribed spheres. It is shown in pure geometry that the
outline projection of a surface of revolution on any plane is
the envelope of the projections of its inscribed spheres ;
that is, the projection may be generated by a circle of vari-
able diameter moving with its centre on the projection of
the axis of the solid.
In illustration, refer to the figure on page 337. The
axis CD of the surface of revolution is parallel to the
vertical plane, and the elevation consists of two circular
arcs, having c'd' as a common chord. The plan of the
surface might be determined thus :
First draw a series of circles inscribed in the elevation ;
then draw the plans of the spheres represented by these
circles ; then draw the envelope, or the curve which
touches all these plans.
XIII
CURVES AND CURVED SURFACES
335
The method of clr wing the plan which is adopted in
the next problem is fuller and more instructive. The curve
HPRG on the surface which projects into the outline in
plan is determined, and a number of points in the plan are
obtained. This gives more definiteness than would be the
case if the method of envelopes alone were used.
336 PRACTICAL SOLID GEOMETRY chap.
282. Problem. A surface of revolution is generated
by a given circular arc CTD, which revolves about its
chord CD. It is required to determine the projections of
the surface when the axis CD is parallel to the vertical
plane and inclined at e to the horizontal plane.
For the elevation draw c'd' making an angle 9 with xy,
and complete the elevation by describing the two arcs on
c'd'. Pet o be the centre of one of these arcs.
To obtain the plan. Suppose a vertical projector to move tan-
gentially round the surface ; then the foot of this perpendicular will
trace the outline of the plan. The projector during its circuit will
touch the surface along a curve, the plan of which will be identical
with the plan of the outline. We shall determine this curve, employing
inscribed spheres for the purpose.
From the centre o draw any radius o'b' , intersecting c'd
in / ; with centre s draw the circle through //, draw b'a per-
pendicular to c'd' , and e's'f parallel to xy. The lines a'b'
and e'f intersect in />'.
Project from c'd' and thus obtain the plan cd, which is
parallel to xy. Project s from s'. With centre s, radius
s'f, describe a circle, and draw the projector from p' to
cut this circle in p, p. Then p, p are points on the
required plan, and p' is the elevation of the point where
the projector touches the surface.
For let S be the centre of an inscribed sphere. Then a'b' is the eleva-
tion of its circle of contact. Let e'f represent a circle on the sphere.
These circles intersect in P. Now the vertical projector through
P touches the surface at P, because it evidently touches the sphere, and
in the immediate neighbourhood of P the two surfaces coincide.
Repeat the above construction for other inscribed spheres.
In particular, draw the radii o't' and o'g' , the latter being
parallel to xy, and obtain the important points R and G,
and by symmetry H.
If a curve be drawn as shown through h', p', r, g', and
other similar points, it will be the elevation of the curve
on the given surface of revolution, such that the outline plan
of the surface is identical with the plan of the curve.
Examples. 1. Work Prob. 282 when c'd' = 3 V', o'f = 2%", 6 = 50 .
Add an auxiliary elevation of the surface on a vertical plane
which makes 45 with xy.
XIII
CURVES AND CURVED SURFACES
337
A surface generated by the revolution of a circular arc about an
external line is 3" long, i|" diameter at each end, and g"
diameter in the middle. Draw the plan when its axis is in-
clined at 45. Draw also the elevation on a vertical plane
making 30 with the plan of the axis.
An annulus is generated by a sphere \\ diameter, whose centre
moves round a circle 2.\" diameter. Draw the plan of the
surface when the annulus rests on a plane inclined at 45.
Hint. The plan is similar in shape to Fig. 123.
338 PRACTICAL SOLID GEOMETRY chap.
283. Miscellaneous Examples.
1. The plane of a circle of 2" diameter is inclined at 50 . A
diameter of this circle is inclined at 30 . Draw the plan of
the circle and also an elevation on a plane parallel to the
inclined diameter. (1893)
*2. Fig. (a). The side elevation of a circular tin canister with the lid
(hinged at A) partly open is given. Draw its plan, and an
elevation on a vertical plane, having xy as ground line. (1885)
3. VTH\s an oblique plane, the traces vt and th making angles of
6i and 71 with xy. A point c, 2.2" from / and 1.7 from xy,
is the plan of the centre of the base of a right circular cone
which has a generator in the horizontal plane and its base in the
given plane VTH. Determine the plan of the cone. (1889)
4. A sphere of radius 1.5" resting on the ground is cut by a plane
inclined at 60", whose horizontal trace touches the plan of the
sphere. Draw the plan of the section. (1877)
5. A vertical cylinder 2" in diameter is cut by a plane inclined at
50 , and having a horizontal trace which touches the plan of
the cylinder and makes 35 with the ground line. Draw the
elevation and development of the curve of section. (1S77)
6. Fig. (b). A horizontal right cylindei lies on a truncated right cone
as shown, the axis of the cylinder passing through that of the cone.
Draw the figure full size, and make a sectional elevation of the
two solids on a vertical plane distant A-" from the axis of the
cone, and making 50 with the axis of the cylinder. (1887)
*7. Fig. {e). v'a'b' is the elevation of a right cone. A circle, whose
centre is 0, is drawn on the elevation, touching v'b' ; and two
tangents to the circle, c'd', c'e', are drawn. Develop the cone,
opening it along the line VB, and determine the developments
of the circle and lines. (1895)
*8. Fig. ((-). Find the real length of the shortest line that can be
drawn on the right cone, whose plan is given, joining the points
whose plans are/, q. Height of cone 3V. (1878)
9. The axis of a cone is inclined at 6o to the H.P., the apex point-
ing upwards and to the right ; the generatrix makes an angle
of 25 with the axis. Determine a section of the cone, the plan
of which will be a circle of 3" diameter. (!897)
*10. Fig. (/). The elevation of a right cone ; and the plan of the
axis are given ; draw the plan of the cone. (Honours, 1880)
*11. Fig. (d). A surface of revolution is shown in elevation ; draw
a. plan on x , y .
12. A cone, base 2.70" diameter, height 2.35", has its axis inclined
at 40. A curve is traced on the cone, which, in development,
would be a circle of 1" radius touching the base of the cone.
Draw the plan of the cone, and of the curve traced on it,
touching the base of the cone at its highest point. (1894)
XIII
CURVES AND CURVED SURFACES
339
CHAPTER XIV
TANGENT PLANES TO SURFACES
284. Nature of the contact of a tangent plane. Let any
three non-collinear points, A, B, C, Fig. (a), near to one
another, be taken on the curved surface of a solid figure,
and suppose a section of the figure to be made by the plane
through A, B, C. If the surface in the neighbourhood be
wholly convex (or wholly concave), like that of a sphere or
the rounded portion of a vase, the section plane will inter-
sect the surface in a closed curve of some kind passing
through the three points ; this is illustrated in Fig. (a).
Consider now any point P on the surface and within this
curve, and suppose the points A, B, and C to approach
indefinitely near to P. Then the plane ABC has a definite,
limiting position, and is called the tangent plane to the sur-
face at P. The tangent plane is said to touch the surface
at P.
Let P be projected on the section plane at /. These
two points are too near together to be distinguished separ-
ately in the figure, and ultimately coincide. Now any line
in the plane ABC through/ will intersect the curve ABC,
and therefore the surface, in two points, shown at D and E
in the figure. Again suppose A, B, C to approach indefinitely
near to P, so that in the limit DE becomes a line through
P in the tangent plane ; hence this line meets the surface in
two consecutive points, that is, N touches the surface or is
tangential to it. We thus see that any line in a tangent
chap, xiv TANGENT PLANES TO SURFACES
341
(*)
plane which passes through the point of contact touches
the surface at the point. Thus we have a series of tangent
lines at any point P on a surface, which all lie in the
tangent plane at P. The tangent plane at P may in fact
be looked on as having been generated by a tangent line
which is rotated about P so as to always touch the surface at
P. The axis of rotation is the normal to the surface at P.
Again, suppose any plane through p, inclined to the
plane ABC, to cut the latter in the line DP, and the surface
in the curve DPP. In the limit this line and curve will
both pass through P, and the line will touch the curve at P.
It is so important that this theorem be well understood
that we present it in several different forms.
Thus, if any plane section of a surface be taken through a
point P on the surface, then the line in which the tangent plane
at P is cut by the section plane is a tangent at P to the curve
of the section.
Or, if a plane touch a surface at P, then the traces of the
plane and the surface 011 any second plane through P also
touch one another at P.
This theorem may be generalised and stated in the
form : If two surfaces touch one another at P, then the traces
of the surfaces on any other surface through P will also touch
o?ie another at P.
As a simple example, suppose a cone to stand with its
circular base on the ground, then the horizontal trace of
any tangent plane will touch the base.
The properties of tangent planes and lines above stated
342
PRACTICAL SOLID GEOMETRY
CHAP.
have been illustrated for the case of a surface wholly convex
or wholly concave in the neighbourhood of the point of
contact P. They are equally true for a surface which at P
is co?ivex in some directions and concave in others, like a horse's
saddle, or the hollow neck of a vase ; but in this case the
surface around /'lies partly on one side of the tangent plane
and partly on the other, and the tangent plane at P cuts
the surface in a curve which has a node at P, Fig. (b), the
two branches having points of inflexion where they cross.
The surface near P is divided into four regions by the two
branches of the curve of intersection. Two opposite regions
lie on one side of the tangent plane, and the other two on
the other side.
285. Surface of revolution with inscribed cone and
sphere. Let QR be any plane curve, and VC any line in
its plane. Let PV and PC be the tangent and normal at
P. With centre C, describe the circle through P.
Now let the figure rotate about
VC as axis. Then the cone and
sphere described by PV and the
circle are tangential to the surface
p? generated by QR, the contact ex-
7\ tending along the circle traced by
P, and represented in the figure by
PP' perpendicular to the axis. The
cone and sphere are said to be in-
scribed in (or they may circum-
scribe) the surface. The tangent
plane to one of the surfaces, at any point in the circle of
contact PP\ also touches the other two.
The surface of a cone or cylinder is generated by the
motion of a straight line, and the tangent plane at any
point of either touches the surface along the line or
generator through the point, and hence touches all in-
scribed spheres.
See Theorems 23 to 48, Appendix II.
XIV
TANGENT PLANES TO SURFACES
343
286. Problem. To determine the traces of the plane
which shall be tangential to a given cone, axis vertical,
vertex V, and shall pass through a given point P on its
surface.
The projections of the cone are shown in the figure.
Draw vb, z/b\ the projections of the generator through
P, and at b draw
the tangent nm \
perpendicular to
vb ; this is the
horizontal trace
of the required
tangent plane.
Through p
draw pc parallel
to nm, and p'c'
parallel to xy.
These are the
projections of a
horizontal line
passing through
P, lying in the
tangent plane,
and terminated at
C by the vertical
plane.
Hence Im
drawn through
c and m is the
required vertical trace.
Examples. 1. Draw the plan and elevation of a cone with its
base on the ground, diameter of base if", height 2^". Draw
the traces of a plane which touches the cone along a generator
whose plan makes 45 with xy.
2. Draw the plan and elevation of a line inclined at 45 to touch the
cone in Ex. 1 at the middle point of the given generator.
3. Determine the traces of a tangent plane as in Ex. 1, but with
the cone inverted so that its vertex rests on the ground.
344 PRACTICAL SOLID GEOMETRY chap.
287. Problem. A given cone, vertex V, rests with its
base on the horizontal plane, determine the traces of a
plane which shall pass through a given external point P
and be tangential to the cone.
The required plane must contain the line VP. Therefore
obtain h, the plan of the horizontal trace of VP.
Draw nhm tangential to the plan of the base of the cone.
Then nm is the horizontal trace of the required plane.
Determine r the elevation of the vertical trace of VP.
Then r'lm is the required vertical trace.
Note. Since two tangents may be drawn from h to the plan of the
base of the cone, there are two tangent planes satisfying the given
conditions.
288. Problem. To determine the traces of the plane
which shall be tangential to a given sphere, centre S, at
a point on its upper surface, the plan p of the point being
given.
First Method. With centre s describe a circle to pass
through p, and obtain a'b' the elevation of this circle ; a
projector from /> to meet a'b' will determine/'. At d and
b' draw tangents to the circle, meeting each other in v and
xy in e',f; th&nv'e'f' is the elevation of a cone which
touches the given sphere along the horizontal circle through
P. The plan of the cone is easily obtained.
Now determine by Prob. 286 the plane LMN to touch
the cone at P ; this will be the required tangent plane.
Second Method. The tangent plane at Pis perpendicular
to SP, hence its traces are perpendicular to sp and s'p' re-
spectively.
To determine these traces draw pc perpendicular to sp
and/V parallel to xy ; then PC is a horizontal line passing
through P, contained by the tangent plane and having C
in its vertical trace. Hence Im drawn through c perpen-
dicular to s'p' will be the vertical trace, and mn perpen-
dicular to sp will be the horizontal trace of the required
tangent plane.
XIV
TANGENT PLANES TO SURFACES
345
X
346 PRACTICAL SOLID GEOMETRY chap.
289. Problem. A given cylinder, axis AB, lies with
a generator on the horizontal plane. It is required to
determine the plane which shall touch the cylinder at a
point whose plan p is given.
Let the rectangle cdefbe the given plan of the cylinder.
First Method. Conceive the tangent plane in position ;
it will touch the cylinder along the generator through P.
To a person looking horizontally in the direction of the
arrow in plan, the cylinder will appear as a circle, and the
tangent plane as a line touching this circle. This view
will now be drawn.
Take x'y perpendicular to ab, and draw the elevation
of the cylinder, that is, the circle with centre a". Obtain
p" by projection from /, and at/" draw the tangent v"o per-
pendicular to d'p" . This is the edge elevation of the tangent
plane, and on drawn through o parallel to ba is the horizontal
trace. The vertical trace hn may be found as in previous
problems, since the horizontal trace and the projections of
a point P in the plane are known.
Second Method. Draw the projections of a sphere in-
scribed in the cylinder and passing through P; then deter-
mine the tangent plane to the sphere at P (Prob. 288).
This will be the required tangent plane (see Art. 285).
Note. If on be nearly parallel to xy, the method of Prob. 211
may be used to determine the vertical trace, as here indicated.
Examples. 1. Draw the plan and elevation of a cone with its
base resting on the ground, diameter of base i|", height 2^",
axis 2" from the vertical plane. A point P is 2" to the right of
the axis of the cone, Ij" above the ground, and i|" from the
vertical plane. Draw the traces of the two planes which pass
through P and touch the cone.
2. Draw the projections of the line from P to touch the cone, so
that the length from P to the point of contact is the least
possible.
3. A sphere i\" diameter has its centre f" above the ground, and
2\" in front of the vertical plane. A point P on the upper
half of the surface is i|" above the ground, and 2" in front of
the vertical plane. Draw the traces of the tangent plane to
the sphere at the point P.
XIV
TANGENT PLANES TO SURFACES
347
Draw the projections of the tangent line to the sphere at the
point P in Ex. 3, such that it is inclined 30 to the ground.
A cylinder iV' diameter touches the ground along a generator
which makes 30 with the vertical plane ; determine the traces
of a plane which touches the cylinder at a point P, ij" above
the ground.
Draw a line through the plan of P in Ex. 6, making an angle
of 6o with xy and 30 with the plan of the axis. If this is the
plan of a line which touches the cylinder at P, draw the eleva-
tion.
348 PRACTICAL SOLID GEOMETRY chap.
290. Problem. To determine the traces of a plane
which shall touch a given cone, vertex V, resting with its
base on the ground, and a given sphere, centre S.
Circumscribe the given sphere by a cone, vertex A, so that
the two cones are similar and similarly placed. The pro-
jections of this circumscribing cone are shown in the figure.
Draw mn an externa/ common tangent to the plans of
the bases of the two cones. This is the horizontal trace of
a required tangent plane.
Determine the projections of C, the vertical trace of the
line VA. Then hn drawn through ;;/ and c is the required
vertical trace.
For, any plane which touches the two cones must touch
the given cone and sphere ; and since the plane contains the
two vertices its vertical trace passes through the vertical
trace of VA.
Note. It will be observed that the plane LAIN is such that the given
sphere and cone lie on the same side of it. There is another such
plane the horizontal trace of which is the other external tangent to the
circles v and a as shown.
There is a second pair of planes each of which passes be-
tween the given sphere and cone. The manner in which the
traces are obtained is exhibited in the same figure, in which
an inverted cone is taken to circumscribe the given sphere,
the base angles of this cone and the given cone being
equal. The trace of the inverted cone on the horizontal
plane is the small circle shown in plan. The internal
common tangents to this circle and the one with centre v
will be the horizontal traces of the two tangent planes, the
vertical traces beinc: found as in Prob. 211.
*&
Examples. 1. Draw the plan and elevation of a cone with its
base on the ground, axis 3" from the vertical plane, diameter
of base 2", height 3". A sphere ij" diameter has its centre
2.y to the right of the axis of the cone, ij" from the vertical
plane, and 1" above the ground. Determine the traces of one
tangent plane to the sphere and cone, and draw the horizontal
traces of all the tangent planes.
XIV
TANGENT PLANES TO SURFACES
349
V
a 4
x
c
77jy
>//l
V
'a
T s;
rt
2. Draw the projections of any line which touches the cone and
sphere in Ex. i.
3. Draw the traces of a plane which shall touch the cone in Ex. I,
page 346, and be J-" from the given point P.
4. Two points A and B are respectively 2" and 1" from each of
the planes of projection. AB is 2f" long. Determine the
traces of a plane through A inclined at 55 to the ground, and
distant f" from B.
5. Determine a sphere which shall contain the point A, Ex. 4, be
distant |" from B, and shall make an angle of 55 with the
vertical plane.
6. Determine a plane which shall touch a sphere, 2" diameter
whose centre is in xv, and shall make 30 with xy.
350 PRACTICAL SOLID GEOMETRY chap.
291. Problem. To determine the traces of a plane
which shall be tangential to two given spheres, centres S
and C, and shall have a given inclination o.
Let the two spheres be each circumscribed by a vertical
cone with base angle equal to the given inclination 6.
These cones may be either upright or inverted, and by
taking the four possible combinations eight common tangent
planes may in general be obtained.
In the figures the horizontal traces only are shown ;
the vertical traces may be found as in the preceding
problems.
In (a) two upright cones are taken.
In (i>) one cone is upright and the other inverted.
In (c) both cones are inverted.
In (d) one cone is inverted and the other upright.
In determining the horizontal traces of the tangent
planes care must be taken to draw the proper common
tangents. Thus in (a) and (c) each tangent plane is such
that the spheres are situated on the same side of it, hence
the external common tangents must be drawn.
On the other hand, in (b) and (d) each tangent plane
passes between the spheres, consequently the internal
common tangents must be drawn to give the traces.
Note. Some of the tangent planes may be coincident or impossible,
and the number of solutions may be anything from eight to none,
according to the data.
Examples. 1. The centres of two spheres which rest on the ground
are 2^" apart ; the centre of one is 2" from the vertical plane,
and that of the other is 1^". The diameters of the spheres
are iiy"and 1" respectively. Determine the traces of a plane
touching the spheres and inclined at 55. Draw the horizontal
traces of all such tangent planes.
2. Two points A, B are respectively 2" and ih" from each plane of
projection and their projectors are 2^" apart. Determine a
plane which shall be inclined at 50, and 1" distant from both
A and B, and which shall lie below A and above B.
3. Determine all the planes which are 1" distant from A and B,
Ex. 2, and which make 50 with the vertical plane.
XTV
TANGENT PLANES TO SURFACES
35i
X
y jl^
x-
c\i
-y
y
352 PRACTICAL SOLID GEOMETRY chap.
292. Problem. To determine the traces of a plane
which shall touch a given cone, axis VA inclined to both
planes of projection, and have a given inclination e.
Let r'v's' and uvw be the projections of the given cone.
Draw the projections of the vertical cone which has its vertex
at f^and its base angle equal to 0.
Draw the projections of any sphere, centre C, inscribed
in the given cone, and also the projections of a cone, vertex
V v similar and similarly placed to the first cone and cir-
cumscribing the sphere.
Then n/n, an external common tangent to the plans of
the bases of the two upright cones, is the horizontal trace of
a tangent plane to these cones, and ml is its vertical trace.
This plane passes through V, touches the sphere C, and so
touches the ajven cone. And it is inclined at 6.
Note I. The inclination 6 cannot be less than that of the least
inclined generator of the given cone.
Note 2. There are, in general, four planes satisfying the given
conditions, two being obtained as shown, and two others from an in-
verted cone (not shown) which circumscribes the sphere C.
Note 3. Instead of the vertical cone through V we might have
taken one circumscribing any second sphere inscribed in the given
cone.
293. Problem. To determine the traces of a plane
which shall touch a given cylinder, axis AB, and have a
given inclination e.
Draw the projections of any two spheres, centres and
C, inscribed in the given cylinder.
Now draw the projections of the two upright cones
which circumscribe the spheres, each cone having a base
angle 9. Determine the plane NML to touch the cones.
Since this plane touches the two cones, it touches both
spheres without passing between, hence it must touch the
cylinder ; and it is inclined at 6.
N te 1. The value of 6 cannot be less than the inclination of the
axis of the cylinder.
XIV
TANGENT PLANKS TO SURFACES
353
X r
2 A
354 PRACTICAL SOLID GEOMETRY chap.
294. Problem. Having given the projections of a
cone, axis VA, and p the plan of a point on its surface, to
determine the traces of a plane which shall pass through
P and touch the cone.
Let r'v's and uvw represent the given cone.
First Method. By Prob. 278 determine/' and also the
projections of T (not shown), the point of contact of PV
and any sphere inscribed in the cone.
The tangent plane to the sphere at the point T is the
required plane ; this may be determined by either of the
methods in Prob. 288.
Second Method. Determine p' as in the preceding
method, and find Q, the horizontal trace of -PV.
If the horizontal trace of the cone be determined, that
is to say, an ellipse whose elevation is r's', the tangent
drawn at q to this ellipse will be the horizontal trace of the
required tangent plane (Art. 284).
But it is unnecessary to actually draw the ellipse. Deter-
mine the foci,/ and f x in plan, as in Prob. 276.
Join qf and qf v and bisect the angle exterior to fqf v
Then the bisector oh touches the elliptic trace at // and is
the horizontal trace of the required tangent plane.
The vertical trace may be determined as in the preceding
problems.
Examples on Problems 292 to 294.
1. The plan and elevation of the axis of a cone make angles of 30
and 45 with xy ; the vertex is 2-J-" above the ground and
ij" in front of the vertical plane. The sphere which is in-
scribed in the cone and rests on the ground is ij" diameter.
Determine the traces of a plane which touches the given
cone and is inclined at 6o to the ground.
2. A sphere 1^-" diameter has its centre l" above the ground and
2-J" in front of the vertical plane. A point A, 2" to the right
of the centre of the sphere, is 1" in front of the vertical plane
and 3" above the ground. Determine the traces of a plane
through A which shall touch the sphere and make an angle of
65 with the ground.
3. The plan and elevation of the axis of a cylinder 2" diameter
XIV
TANGENT PLANES TO SURFACES
355
4.
make angles of 25 and 40 with xy. Determine the traces
of a plane which shall touch the cylinder and have an inclina-
tion of 65. Draw the horizontal traces of all the planes which
satisfy the given conditions. What are the least and greatest
possible inclinations of the planes which touch this cylinder?
Take the cone Ex. I. Select a point p i|" to the left of the plan
of the vertex, and 2" from xy. This is the plan of a point
on the upper half of the cone. Determine the traces of the plane
which touches the cone at P.
Determine a plane which shall touch the cylinder of Ex. 3 along
a line whose plan is A-" distant from the plan of the axis.
Determine all the planes which touch the cylinder of Ex. 3 and
make 6o with the vertical plane.
356 PRACTICAL SOLID GEOMETRY chap.
295. Problem. To determine a plane which shall
contain a given line AB and touch a given sphere C.
First Method.- It is obvious that the required plane will
touch any cone circumscribing the given sphere and having
its vertex in AB. Let the particular cone with vertex Vbe
taken, where v is determined by drawing cv parallel to xy.
Such a cone has its axis VC parallel to the vertical plane.
Project v from v, and draw the tangents v'r, v's' ; then
r'v's' is the elevation of the cone.
Draw the elevation of the vertical cyclinder which
circumscribes the given sphere.
The cylinder and cone intersect in two ellipses whose
edge elevations are the lines e'd' and/'^"' respectively. The
plan of each ellipse is the circle centre c, since both ellipses
are on the cylinder (see Art. 273).
Produce e'd' to meet a'b' in ?i and xy in /; then TLM is
an inclined plane which cuts the cone in an ellipse, the eleva-
tion of which is e'd'.
Conceive the tangent plane in position ; it intersects the
plane TLM in a line which passes through iVand touches
the ellipse whose elevation is e'd' (Art. 284). The plan of
this line is nz (or nk), touching the circle which is the pro-
jection of the ellipse.
Therefore the required tangent plane contains the lines
NZ (or NK) and AB, the horizontal traces of which are
h and 0, in plan. Hence ho is the horizontal trace of one
tangent plane satisfying the conditions of the problem.
The vertical trace is found as in previous problems.
Note 1. The point of contact of tangent plane and sphere is the
point P where VZ meets the circle of contact of cone and sphere.
Note 2. If the tangent NK be taken instead of NZ another
tangent plane is found. Thus there are two solutions.
Note 3. The inclined plane through fg' might have been taken
instead of the one through e'd! , but this would not have met AB
within convenient limits.
Note 4. If the position of AB be such that it is inconvenient to
determine V, a new elevation of the line and sphere may be drawn and
the same method applied.
XIV
TANGENT PLANKS TO SURFACES
357
tf\ - r '
358 PRACTICAL SOLID GEOMETRY chap.
Second Method. A modification of the first method is
illustrated in the figure on page 359. The position of AB
is altered but the same letters are used.
Draw the elevation of the cone, vertex V, as before.
Draw the traces ///;/ of the inclined plane which contains
the circle of contact of cone and sphere. The given line
intersects this plane in the point N.
Suppose the tangent plane in position. It will intersect
the plane TLM in a line which is a tangent from N to the
circle whose elevation is e'd'.
An auxiliary plan of the circle and the tzvo tangents
from ./Vis drawn on x 1 y 1 taken parallel to //. Here ij = qc.
The required plane contains NK or (JVZ) and AB ; its
horizontal trace therefore passes through and // as before.
Note 1. Since A' is in the tangent plane, the latter may be deter-
mined by finding the traces of any two convenient lines through A'
which meet AB.
Note 2. The tangent plane touches the cone along the generator
KV, hence it touches the sphere at a point P in KV. When P has
been found the problem is reduced to Prob. 288.
Third Method. Take a cone with its vertex at any
convenient point in the given line and circumscribing the
given sphere.
Find the horizontal trace of this cone, and draw the two
tangents to the trace from the horizontal trace of the given
line. These two tangents are the horizontal traces of the
required tangent planes.
This method is the more straightforward to apply, but
requires the setting out of the curved trace, which is a conic
section. The other methods require only circles and straight
lines to be drawn.
Example. Draw a line inclined at 6o to xy, in which take a
point a 1" from xy, and another point b 2^" from xy ; this
is the plan of a line AB, whose ends A and B are 3" and 1"
above the ground. Describe a circle Ij" diameter, with its
centre 2" from both xy and ab ; this is the plan of a sphere
whose centre is 2" above the ground. Determine the traces of a
plane containing the line AB and touching the sphere.
XIV
TANGENT PLANES TO SURFACES
359
360 PRACTICAL SOLID GEOMETRY chap.
296. Problem. To determine the traces of a plane
which shall touch a given cone, axis VA, and pass through
a given external point P.
First Method. This problem may be at once reduced to
the preceding one, since the required plane must contain
the line joining the vertex of the cone to the given point,
and touch any sphere inscribed in the cone.
There will be two planes fulfilling the required condi-
tions.
Second Method. Let t'r's and vuw be the projections
of the given cone, and p\ p those of the point.
The required plane contains PV, hence its horizontal
trace passes through the horizontal trace of PV ; therefore
determine h, the plan of the horizontal trace of PV.
The horizontal trace of the tangent plane must also
touch the horizontal trace of the given cone, that is, an
ellipse whose elevation is //.
Determine the foci/j and /in plan, and the major axis
of this ellipse, as in Prob. 276.
On the major axis as diameter describe a semicircle ;
and on hf x as diameter describe another semicircle, the two
intersecting in n.
Then /// drawn through n is a tangent to the ellipse,
and is therefore the horizontal trace of a tangent plane.
And It is the vertical trace, It being drawn through //, the
elevation of the vertical trace of PV.
Note. Two tangents can be drawn from h to the horizontal trace
of the cone, hence there are two tangent planes which fulfil the
required conditions.
Examples. 1. Determine the traces of a plane which shall touch
the cone of Ex. 1, page 354, and pass through a point P, 1" to
the right of the vertex, 2" in front of the vertical plane, and 1"
above the ground.
2. The axis VA of a cone, vertical angle 25 , coincides with xy. A
point P is 1 \" distant from both planes of projection, and the
projector//' contains v, the projection of the vertex. Deter-
mine a plane through P to touch the cone. Find also a plane
which touches the cone and is inclined at 45 .
XIV
TANGENT PLANES TO SURFACES
361
362 PRACTICAL SOLID GEOMETRY chap.
297. Problem. The projections of a cylinder, axis
AB, being given, and p the plan of a point P on the
surface of the cylinder, it is required to determine the
traces of a plane which shall pass through P and touch
the cylinder.
First Method. The required plane will touch the
cylinder along the generator through P.
Determine/' by the method of Prob. 279, and draw the
projections of the generator PD.
This generator touches any inscribed sphere, centre C,
at T, the projections of which may be determined at the
same time that/' is found.
The plane which touches the sphere at T is the required
tangent plane, and may be found as in Prob. 288.
Second Method. After having determined the projections
of the generator PD, the horizontal trace nm of the tangent
plane may be determined by first obtaining Q, the hori-
zontal trace of PP>, and then drawing a tangent at q
to the ellipse whose elevation is r s .
This is done in precisely the same manner as was ex-
plained in the second method of Prob. 294, Imn being
the traces of the required plane.
298. Problem. To find the traces of a plane which
shall touch two given spheres and pass through a given
point. (No figure.)
This may be reduced to Prob. 296 by circumscribing
both spheres by the same cone. The plane which passes
through the given point and touches this cone must satisfy
the given conditions.
There are two such cones, one of which has its vertex
between the spheres. Thus there are in general four
tangent planes, two being found from each cone.
299. Problem. To determine the traces of a plane
which shall be tangential to three given spheres.
Take the spheres in pairs, and determine the vertices
XIV
T ANCIENT PLANES TO SURFACES
363
V, IV of two cones which envelop any two of the three
pairs ; then determine a plane which contains VW and
touches any of the spheres.
The projections of V and W are found by drawing the
common tangents to the projections of the spheres, and
the solution of the problem is then the same as that of
Prob. 295.
364 PRACTICAL SOLID GEOMETRY chap.
390. Problem. To determine the traces of a plane
which touches a given surface of revolution, axis vertical,
at a given point P on its surface.
First Method. The surface considered is the same as
that in Prob. 282, d being the centre of the arc c'b'd'.
One projection of /'being given, the other can be found by
means of a horizontal section through the point, viz. dp b'
in elevation and the dotted circle through/ in plan.
Join db\ meeting c'd' in s . With / as centre and s'b' as
radius describe a circle. This is the elevation of the in-
scribed sphere which touches the given surface in the circle
passing through P.
And LMN, the tangent plane at P to this sphere, found
as in Prob. 288, will also touch the given surface at P
(Art. 285).
Second Method. Draw the projections of a vertical cone
touching the surface in a circle, of which db' drawn through
p' is the elevation. Then P lies on the cone, and the plane
touching the cone at P also touches the given surface.
This tangent plane may be found as in Prob. 286.
Examples on Problems 297 to 390.
1. The plan and elevation of the axis of a cylinder 2" diameter
make angle of 30 and 40 with xy. Determine the traces
of a plane which touches the cylinder at a point whose plan
is |" from the plan of the axis.
2. Determine the traces of a plane which shall touch the two spheres
of Ex. 1, p. 350, and pass through a point i^" vertically over
the middle point of the line joining the centres of the spheres.
3. Three spheres of 1.5", 1.0", and .4" diameters rest on the
ground in mutual contact, and a fourth sphere .8" diameter
rests on the three. Draw the plan of the group and determine
the tetrahedron which envelops them, each face touching three
spheres.
4. Draw an equilateral triangle abc of 2" side, and with a, />, c as
centres describe circles of 1", ", and ^" radii respectively.
Take these circles as the plans of spheres which rest on the
ground, and determine a plane to touch all the spheres.
Hint. The horizontal trace of the tangent plane is found
at once since the vertices of all the enveloping cones are on
XIV
TANGENT I'LANES TO SURFACES
365
the ground. The inclination of the plane is then readily deter-
mined by taking an elevation on a vertical plane perpendicular
to the horizontal trace.
5. Take the surface of revolution of Ex. 2, p. 337, with its axis
vertical and one end on the ground, and determine a plane
which shall touch the surface at a point 2|" above the ground,
the line joining the plan of the point to the plan of the axis
making 45 with xy.
6. A surface of revolution is shown in elevation at (f) on p. 485.
Copy the figure double size and draw the plan. Determine a
plane which shall touch this surface and make angles of 6o and
40 with the horizontal and vertical planes of projection.
7. In Ex. 6 determine the plan and elevation of the section of the
capstan by the tangent plane, and show that the section has a
node at the point of contact.
*66 PRACTICAL SOLID GEOMETRY chap.
301. Miscellaneous Examples.
*1. Eig. (a). The plan of an inverted right cone with its vertex
on the horizontal plane is given. The height of the plane of
the base is 25. Determine the scale of slope of a plane con-
taining the given point .Panel tangential to the cone. Unit o. 1".
(1887)
*2. Fig. (/;). The plans of two spheres are given and also the plan
of a point P. Determine a plane touching both spheres, and
passing through P. ( x 893)
*3. Fig. (c). The given sphere rests on the horizontal plane. Deter-
mine the scales of slope of planes touching it and containing the
given line AB. Unit o. 1". (1884)
*4. Fig. (d). ad is the plan and a the trace of a line inclined at 50
to the horizontal plane. AB is the axis of a right cylinder of
1.5" diameter and 1. 5" in length, its lower base resting on the
horizontal plane. Draw the plan of the cylinder, and the hori-
zontal trace of a plane tangent to it and inclined at 70 to the
horizontal plane. ( 1S95)
*5. Fig. (e). 9 ^ 8 is the axis of a right cylinder of 1 5 units diameter,
and p.22 is the centre of a sphere of the same diameter. Draw
a plane passing over the cylinder and under the sphere, and
touching the surfaces of both. Show the point of contact be-
tween the plane and the sphere. Unit = o. 1". ( X S96)
*6. Fig. (_/). Determine a plane inclined at 65, making 70 with
the given plane, and |" distant from the given point. Unit =
0.1". (Honours, 1889)
*7. Fig. (g). Two spheres are given, the index of the centre of the
smaller being 11, and the lowest point of the larger being level
with the highest point of the smaller. Draw a plane inclined at
70 to the horizontal plane, and touching both the spheres.
Unit = 0.1". (Honours, 1891)
8. A sphere 1.75" in diameter touches both planes of projection.
Determine the traces of a plane touching the sphere, and in-
clined at 6o and 50 to the horizontal and vertical planes
respectively. (1878)
9. Three planes are mutually perpendicular, and each touches a
sphere 1" diameter, which rests on the ground ; two of the
planes are inclined respectively at 35 and 70 . Draw the
horizontal traces of the three planes, the plan of their intersec-
tions, and find the inclination of the third plane.
10. A right cone, 2^" high, rests with its base (of ij" radius) on
the H.P. A sphere of #'' radius touches it externally. Draw
the true shape of the section of the cone, made by a plane pass-
ing through the vertex of the cone, inclined at 75 to the H.P.
and passing under, and touching the sphere. (1S98)
XIV
TANGENT l'LANES TO SURFACES
367
CHAPTER XV
SURFACES IN CONTACT
302. General remarks. Two surfaces which touch will
have a common tangent plane and a common normal at the
point of contact. The normal at any point on a sphere is
the line joining the point to the centre ; and at any point
P on the surface of a circular cone or cylinder, or any
solid of revolution, is the line joining P to the centre C of
the inscribed sphere which contains P. See notes to Prob.
278. Any surface which touches the sphere at P also
touches the surface of revolution at the same point. Thus
problems on the contact of cones and cylinders may often
be reduced to those on the contact of spheres, and thereby
simplified. See Theorems 23 to 48, Appendix II.
303. Problem. To determine the projections of three
spheres, centres A, B, C, of given radii, which shall rest
on the ground in mutual contact.
Assume that the spheres with centres A and B are
placed so that AB is parallel to the vertical plane.
Describe circles, centres a and b', to touch xy and
each other, their radii being those of the spheres A and B.
These circles are the elevations of two of the spheres.
Draw their plans with ab parallel to xy.
Describe a circle (centre c ') to touch xy and the
circle with centre //, the radius being that of the third
sphere. Describe an equal circle to touch the one with a
CHAP. XV
SURFACES IN CONTACT
369
M^7
as centre. Then inn and or are the lengths of the plans
of BC and AC. Hence c, the plan of the centre of the
third sphere, is a point of intersection of two arcs drawn
one with centre a, radius or, and the other with centre b,
radius mn.
Draw c-[c parallel to xy to meet a projector from c in c .
This gives the elevation of C, and the projections of the
third sphere may now be drawn.
The point of contact for each pair of spheres is indi-
cated ; thus e is the intersection of b'c with a line through
e x ' parallel to xy ; and e projected from c is in be.
Example. Draw the projections of three spheres which touch
each other and rest on the ground, the diameters being 2j",
ij", 1". Show the projections of the points of contact.
2 B
37Q PRACTICAL SOLID GEOMETRY chap.
304. Problem. To determine the projections of a
sphere of given radius which shall touch (externally) a
given sphere, centre S, at a given point P.
The centre of the required sphere is on SJ 3 produced,
therefore produce sp and s'p' .
-Through p' draw the horizontal line a'b ' .
Join s'b' and produce to c^, making c/b' equal to the
given radius. With centre C-[ describe the circle through //.
Draw CyC parallel to xy to meet s'p' produced in c,
and project c on sp produced. With centres c and c,
radius c-[b\ describe circles ; these are the projections of
the required sphere.
This construction is suggested by conceiving the re-
quired sphere to roll on the given one until SC is parallel
to the vertical plane, the point of contact remaining on
the horizontal circle through P; their elevations while in
this position can readily be drawn. The line c'c^ is the
elevation of the path of C during this motion, and p'b' that
of the point of contact.
305. Problem.; To determine a sphere which shall
rest on the ground and touch at a given point P a given
sphere, centre S, which also rests on the ground.
Through /' draw p'b' parallel to xy. Join s'b', and on
s'b' produced, determine /, the centre of a circle which
touches xy and the circle centre / (see Prob. 73)
Draw c/c parallel to xy to meet s'p' produced in c ; then
c is the elevation of the centre of the required sphere, the
plan c being obtained by projection.
With centres c and c, radius c/b', describe circles ;
these are the projections of the required sphere.
Examples. 1. A sphere 2'' diameter rests on the ground, with its
centre C 1^" from the vertical plane. Determine a sphere,
1 1" diameter, to touch the given one at a point P, whose plan
p is 2^" from xy, and i" to the right of c.
2. A sphere 2|" diameter, centre C, rests on the ground.
Determine a sphere which shall rest on the ground, and
touch the given sphere at a point whose elevation is 1"
above xy and Jf" to the right of c' .
XV
SURFACES IN CONTACT
371
372 PRACTICAL SOLID GEOMETRY chap.
306. Problem. To determine the projections of a
sphere of given radius which shall touch (externally) a
given vertical cone, at a given point P on its surface.
Through p' draw al> parallel to xy, and draw b's at
right angles to v'd' to intersect the elevation of the axis of the
cone in /. If a circle (not shown) were described with s
as centre and s'b' as radius, it would be the elevation of a
sphere inscribed in the cone, and touching it in the
circle whose elevation is db' .
The required sphere must touch this sphere at P,
hence its projection may be found as in Prob. 304, and is
shown in the figure.
307. Problem. To determine the projections of a
sphere of given radius which shall touch a given inclined
cylinder, axis DE, at a point whose plan p is given.
Determine the elevation of /'(see Prob. 278, note 1), and
also the projections of the sphere, centre S, which passes
through P.a.nd is inscribed in the cylinder.
The required sphere must touch this sphere at P; its
projections may therefore be found as in Prob. 304, and
the construction is shown in the figure.
308. Problem. To determine the projections of a
sphere which shall rest on the ground and touch a given
inclined cone at a given point P (no figure).
Obtain the projections of the sphere which is inscribed
in the given cone and passes through P (see Prob. 278,
note 1). The required sphere must touch this sphere at
P, and its projections may be found as in Prob. 304.
Examples. 1. A cone rests with its base on the ground ; diameter
of base 2", height 3". Draw the plan and elevation of a
sphere, i#" diameter, which shall touch the given cone at a
point 2" from the vertex ; the elevation of this point being -j?"
to the right of that of the axis of the cone.
2. The plan and elevation of the axis of a cylinder, 2"
diameter, make angles of 45 and 30 with xy. Determine
a sphere, 1^" diameter, which shall touch the given cylinder
at a point whose plan is " from the plan of the axis of the
cylinder.
XV
SURFACES I\ CONTACT
373
\W
374 PRACTICAL SOLID GEOMETRY chap.
309. Problem. The plans of a sphere, centre C, and of
a line AB which touches the sphere are given, the indices
of c and a being attached ; determine and index the plan
of the point of contact.
Regard ab as the plan of a vertical plane ; the plane
will intersect the sphere in a circle, diameter de, to which
AB must be a tangent.
Take xy parallel to ab, and project the elevations of
A and the circle centre O, making md = 5 units. From a
draw the tangent a'p' ; then this is the elevation of AB, and
P is the required point of contact. Measure p'n and pro-
ject and index/ accordingly.
310. Problem. The circles, centres v and c, are the
plans of a cone and sphere which rest on the ground,
the height of V being given. Determine the plan of a
cylinder of given radius which rests with a generator on
the ground and touches the given cone and sphere.
On xy, parallel to vc, draw the elevations of cone and
sphere.
Describe a circle, centre a^, radius equal to that of the
required cylinder, to touch xy and the elevation of the
sphere. Describe an equal circle to touch xy and e'v .
These circles may be regarded as end elevations of the
cylinder, and the plan of the axis of the required cylinder
is distant from c and v lengths equal to aJn and b^m.
Hence, with centres c and v, radii a/n and ml\, describe
arcs and draw ab tangential to these arcs. This is the plan
of the axis of the required cylinder, and the outline of the
plan can now be at once drawn.
Examples. 1. A sphere, 2" diameter, centre C, rests on the ground.
A line a_.b, J" from c, is the plan of a tangent to the sphere at
B ; _ 5 6=2j". Determine and index the plan b. Unit o. 1".
2. A cone rests with its hase on the ground, diameter of base if",
height 2^". A sphere ij" diameter also rests on the ground,
its centre being i^" from the axis of the cone. Determine a
plan of a cylinder, 1" diameter, which rests with a generator on
the ground and touches the given cone and sphere, and show
the points of contact.
XV
SURFACES IN CONTACT
375
376 PRACTICAL SOLID GEOMETRY chai\
311. Problem. To determine the projections of two
cones which rest on the horizontal plane in line contact
with one another, the vertical angles being e and 9.
One axis VZ is taken parallel to the vertical plane.
Since the cones must touch along a common generator,
and also lie on the ground, their vertices must coincide.
Draw the isosceles triangles v't'f and v't'g-^, with the
vertical angles at v equal to 6 and <. These are the
elevations of the two cones in line contact, with their axes
parallel to the vertical plane.
Take any point s in the axis v'z and draw s'a'^ per-
pendicular to v't' . Then ^ and C x are the centres of two
inscribed spheres touching each other at A v
Project vsz parallel to xy ; with centre s, radius s'a' v
describe a circle ; the tangents from v to this circle form
the plan of the larger cone.
Conceive the cone V\V to roll over the other until it
lies on the ground ; its projections when in this position
must now be found.
Draw 1m parallel to, and distant C-[a^ from xy ; also
draw c/c perpendicular to v'z , meeting lin in c . With
centre c describe a circle to touch xy, and from v draw
the upper tangent. The elevation of the smaller cone
is thus found.
With centre s', radius s'c^, describe a circle cutting
hn in e ; draw s'r at right angles to hn. With centre s,
radius re, describe an arc to cut a projector from c in
c ; then c is the plan of C. The plan of the smaller cone
is completed by drawing tangents from v to the plan of
the sphere, centre C.
To explain the construction. As the cone VIV rolls over the
other, the triangle SCV turns about SV until C comes into the plane
LM; c^c is the edge elevation of the circular path of C. During this
motion C remains on the surface of a sphere, centre S, radius s'c^ ;
hence the plan c is found.
Note. Draw a-[ci parallel to c^c' , and project a on sc. Then A
is the point of contact of the spheres, and VA the line of contact of
the cones.
XV
SURFACES J\ CONTACT
377
Example. Determine the projections of two cones which rest on
the horizontal plane in line contact with one another, the ver-
tical angles being 50 and 25 respectively.
378 PRACTICAL SOLID GEOMETRY chap.
312. Problem. The scales of slope of two planes are
given, to determine the plan of a cone, vertical angle
e, which lies between the planes so as to touch them.
The figure to the left shows an elevation of the cone
with a sphere of any radius z inscribed in it.
Conceive the cone lying between, and in contact with,
the given planes. The vertex Fmust be at some point on
their intersection, and C, the centre of the inscribed sphere,
will be on the intersection of two planes respectively
parallel to the given planes and distant z therefrom ; C will
also be situated on the surface of a sphere, centre V, and
radius V-[c^.
Determine r's and lik\ edge views of the given planes,
and draw lm' and f'g respectively parallel to and distant
z from them.
Determine a l> 10 , the plan of the intersection of the given
planes, and also d Q e 1Q , the plan of the intersection of the
second pair of planes (see Prob. 244).
In a & 1Q select any point (in AB\ say z> 5 , as the plan of
the vertex, and with v b as centre and radius v-[ c^ describe
a circle ; this is the plan of a sphere on the surface of which
C must be situated. But C is also in DE.
Therefore by a construction similar to that in Prob. 309
determine the intersection of DE and this sphere ; the plan
of one point of intersection is c n - s . With centre c and
radius z describe a circle and draw the tangents to it from
v 5 ; the required plan of the cone is thus determined.
Examples. 1. Draw oc and od including 120". On oc take
or = |" and os = 2^". On od take oc = h", oc 2". Regard r Q s 15
and f-5e 10 as the scales of slope of two planes. Determine the
plan of a cone, vertical angle 45, which lies between the planes
so as to touch them. Unit 0. 1".
2. Determine the plan and elevation of a cone of indefinite length,
vertical angle 6o, to which the planes of projection are both
tangential. Show the lines of contact.
3. The traces vt, th of a plane make 40 and 60 with xy. Find
a right-angled cone which touches this plane and the ground.
XV
SURFACES IN CONTACT
379
General Examples. 1. Three spheres of diameters 2", ih", and
1" rest on the ground in mutual contact, antl a fourth sphere, |"
diameter, rests on the three. Draw the plan of the group.
Draw also the plan of the triangular pyramid which circum-
scribes the spheres, showing the three points of contact on
each face of the pyramid.
2. A line AB, 3" long, has its ends in the planes of projection, and
is inclined at 6o to A'Kand 40 to the ground. Determine
a cylinder having XY as its axis, which shall touch the
line AB.
3. Determine a sphere of 2" radius which shall have its centre in
A'Kand touch the line AB of Ex. 2.
4. A sphere, 3" diameter, has its centre C in XY; a. second sphere,
ivr" diameter, centre S, touches both planes of projection ;
SC=2h"- Determine a sphere which shall rest on the ground
and touch the cone circumscribing the given spheres at a point
distant f " from both planes of projection.
380 PRACTICAL SOLID GEOMETRY chap.
313. Problem. A given cone, axis VZ, lies on the
ground ; to determine the projections of a cylinder of
given radius which shall touch the cone externally at a
point whose plan p is given, the direction, mn, of the
plan of the axis of the cylinder being also given.
Suppose the required cylinder in position touching the
given cone at P, then
(1) There will be two spheres inscribed one in each so
as also to touch one another at P ; call their centres C
and S.
(2) The generators of cylinder and cone which pass
through P must both lie in the tangent plane common to
the cylinder and cone, or the inscribed spheres, at P.
Determine, by Prob. 278, the elevation of P and the
projections of the sphere, centre S, which passes through P
and is inscribed in the cone.
Also by Prob. 278 determine the projections of a
sphere, centre C, which touches the given cone at P,
and has a radius equal to that of the required cylinder.
The required cylinder must circumscribe this sphere,
hence its plan is obtained by drawing those tangents to
the circle, centre c, which are parallel to mn.
Draw vt perpendicular to sp. Then vt is the horizontal
trace of the tangent plane referred to in (2) above.
Draw ph parallel to mn, and project Ji ; join p'h' .
Then p'h' is the elevation of a generator PH. Hence
the elevation of the cylinder is obtained by drawing those
tangents to the circle, centre c , which are parallel to p'h'.
Note. If it were required that the cylinder should touch the
cone along a generator, so as to have line instead of point contact, the
tangents to the circle c, giving the plan, would be drawn parallel to/?'.
Example. Draw two lines including an angle of 45 ; these form
the outline of the plan of a cone which touches the ground along
a generator. Determine the projections of a cylinder ij"
diameter, the plan of whose axis makes 30 with the plan of the
axis of the cone, which shall touch the cone at a point whose
plan/ is 1" and \" respectively from the first two lines.
XV
SURFACES IN CONTACT
38i
General Examples. 1. A sphere 2\" diameter has its centre C
in A'F. A point A is distant i" and 2" from the vertical and
horizontal planes of projection and 3" from C. Determine a
sphere, centre A, which shall touch the given sphere.
2. A cone, vertex V, vertical angle 6o, has its axis along XY. A
point A is distant 1" and i|" from the horizontal and vertical
planes, and 2" from V. Determine a normal from A to the
cone, and a sphere, centre A, which touches the cone.
3. Draw the projections of any point A, and of any cone with its
axis inclined to both planes of projection. Determine the two
normals from A to the cone, and the two spheres, centre A,
which touch the cone.
4. A cylinder 2" diameter and a cone, vertical angle 60, lie on the
ground in contact with their axes at right angles. Draw their
plans and determine a sphere of \\" radius which shall rest on
the ground and touch both.
382 PRACTICAL SOLID GEOMETRY chap.
314. Miscellaneous Examples.
1. A sphere of 2^" diameter touches both planes of projection. A
second sphere, diameter i|", touches the first sphere, and has
its centre in the ground line. Draw the projections of the two
spheres. (1889)
2. Two spheres, diameters 2.25" and 1", rest on the horizontal
plane touching each other. Draw the plan of the complete
locus of the centre of a sphere of 1.4" diameter, touching both
spheres. (1888)
3. A right circular cone, the vertical angle of which is 35 , rests
with a generator in the horizontal plane. A sphere of i|"
radius also rests on the horizontal plane, and touches the cone
in a point 2|" from the vertex. Draw the plan of the two
solids. (1892)
*4. Draw a sphere, I ' radius, resting on the horizontal plane, and
touching the two given planes. Determine the points of con-
tact. Unito.i". (1885)
*5. Two planes are given by their traces aob, cod. Draw the pro-
jections of a sphere 2" in diameter, touching the given planes,
and having its centre in the horizontal plane. (1887)
*6. Draw the plan of any sphere such that the given line AB is
tangent to it, and that the centre of the sphere is in the line
CD. Unit = 0.1". (1894)
*7. The plans of a right cylinder and of a sphere are given. A right
cone, diameter of base 2^-", height 3^", stands on the horizontal
plane and touches both cylinder and sphere. Draw its plan
and show the points of contact. (1886)
*8. Determine the centre and radius of a sphere to which the two
given lines AB, CD shall be tangent. Unit = o.i". (1881)
*9. A plane is given by its scale of slope, and a line AB by its
figured plan. Determine the centre of a sphere of i|-" radius,
touching the given plane and line, and resting on the horizontal
plane. Unit o. 1". (1890)
*10. A right cone is lying on its side upon the horizontal plane, b is
its vertex, and ab is the plan of its axis, which is inclined at
25 . The point c is the plan of the centre of a sphere, which
also rests upon the horizontal plane, and which is in contact
with the cone. Complete the plan. (1882)
XV
SURFACES IN CONTACT
383
Co/iy /he fiaures double siz&.
10 20 30 40
I I I h
-5
e
-s
-e
9
CL
Ik
c
15
CHAPTER XVI
INTERSECTIONS OF SURFACES, OR INTERPENETRATIONS
OF SOLIDS
315. The general problem and its solution. When two
solids penetrate each other, the nature of the line of inter-
section of their surfaces depends upon that of each of the
surfaces. Thus if both surfaces consist of a series of plane
faces, they will intersect each other in a series of straight
lines ; or if one or both of the surfaces be curved, the
intersection will consist of one or more curves ; these curves
may be plane, but generally are tortuous curves ; that is,
such as cannot be contained by a plane.
One solid may completely penetrate the other, entering
at a closed curve or zigzag line where their surfaces inter-
sect, and emerging at a second closed curve or gauche
polygon. Or the interpenetration may be only partial, in
which case the line of intersection generally consists of
only one closed figure. For the case intermediate between
these two, the surfaces of the solids touch in one or more
points ; here we must expect to find a node at a point of
contact, that is, two branches of the curve of intersection
may cross at the point.
Method of sections. The method most commonly
adopted for determining points on the intersection of the
surfaces of two solids, or of any two surfaces, is that known
as the method of sections. The solids are supposed to be
cut by a series of section surfaces, generally plane, but
sometimes spherical or otherwise curved. The shapes of
chap, xvi INTERSECTIONS OF SURFACES 385
the sections are drawn in plan and elevation, and the series
of points where these intersect are thus determined. These
points are common to the surfaces of the two solids ; that
is, are points in the required intersection. The section
surfaces are chosen, if possible, so that the projections of
the sections shall always be straight lines or circles both in
plan and elevation, these being the only two forms that can
be drawn without trouble.
It would at first appear that the greater the number of
sections taken, and hence of points determined, the greater
would be the accuracy with which the curves of intersection
could be drawn. This, however, would only be true if each
point could be located with absolute accuracy, which is not
possible. So the greater the number of points the more
marked will be the result of any slight error in determining
any one. To obtain the best results, comparatively few
section surfaces should be taken, but their positions
should be very carefully and judiciously chosen.
In almost all cases there will be certain important .
special points on the line of intersection whose projections
ought to be found. Such, for instance, are all points which,
in the projections, fall on the outlines of the figures. The
projected curve of intersection generally touches the outline
at such points, and these points often separate a visible from
an invisible portion of the intersection. The general
method of procedure should be first to select those section
surfaces which give us the important special points. These
may be sufficient to enable us to plot the whole curve.
But should there be any wide intervals, one or two extra
sections may be taken to fill up the gaps.
In the. problems and figures which follow, the positions
of the important sections are generally indicated. A
section is projected in detail, and the corresponding points
of the curve of intersection determined. The projections
of the other sections are omitted to avoid confusing the
figures, though these may have had to be drawn to enable
us to give the complete intersections.
2 c
386 PRACTICAL SOLID GEOMETRY chap.
316. Problem. To determine the plan and elevation
of the section of the given sphere, centre 0, by the given
oblique plane LMN.
Here one of the two intersecting surfaces is a plane.
In determining points on the curves, it is convenient to
employ a series of sections by horizontal planes, for these
project into straight lines and circles in plan, and of course
into straight lines in elevation.
One such section plane is drawn in edge elevation at p'q .
It cuts the sphere in a circle of diameter s x s v and the plan
of this circle is drawn with centre o. And it cuts the plane
LMNrn. the line whose plan is pq, determined by projecting
from/' to/ and then drawing/^ parallel to mn.
The plans of the sections of sphere and plane intersect
in s, s ; and the elevations /, / are found by projecting
from s, s on to p'q' .
The points S, S are common to the section plane PQ,
the plane LMN, and the sphere, and are therefore points
on the required intersections of the two latter.
This illustrates the method of determining points by
any section plane ; we must now select those sections which
give rise to the most important points on the curves.
First as to the upper and lower limits of the curve.
Draw ok perpendicular to mn; let this be the plan of a line
in the plane LAIN; the line will evidently intersect the
required section in its highest and lowest points. To
determine the levels of these we have drawn the elevation
A x A x of the line, supposing it to have been turned into a
position parallel to the vertical plane about a vertical axis
through O. The points A v A v where this elevation cuts
the elevation of the sphere, give the highest and lowest
levels for the section planes. The two sections at these
levels give the two points A, A ; at a', a the required
curve is horizontal ; at a, a the tangents are parallel to mn.
Next as to the points on the outlines. A horizontal
plane through the centre of the sphere will cut the latter
in a great circle which projects into the outline in plan ;
XVI
INTERSECTIONS OF SURFACES
3*7
this section gives the two points B, B on the curve, whose
plans b, b are on the outline plan of the sphere. Observe
that the plan of the curve touches the outline at the points
b, b where the two meet ; note also that the points b, b
separate the full and dotted parts of the curve, which repre-
sent the visible and hidden portions.
To obtain the points on the outline in elevation, we
must take a section which projects into this outline. This
is given by a plane through O parallel to the vertical plane.
The elevation of the line where this plane cuts the plane
388 PRACTICAL SOLID GEOMETRY chap.
LMJY'is shown ; it intersects the outline in c, c . Thus the
points C, C are given by this vertical section plane.
Again observe that the curve and outline touch at /, /,
and that the curve changes from a full to a dotted one or
vice versa when it passes a point on the outline.
The eight points S, A, B, C thus determined are almost
sufficient to enable the curves to be plotted, especially as the
latter are ellipses, to the form of which the eye is well
accustomed. Otherwise one or two intermediate section
planes will give the required additional points.
Examples. 1. Copy Fig. 316 double size, and work the problem.
2. A cone stands upright on the ground, draw the projections of its
section by an oblique plane. Diameter of base 2.2"; height
2.6". Horizontal trace of plane touches base, and both traces
make 50 with xy.
3. A square pyramid stands upright on the ground, obtain the
projections of its section by an oblique plane. Height 2.5",
side of base 1.8", making 30 with xy. Horizontal trace of
plane 1.2" from plan of axis, both traces making 45 with xy.
Hint. Take section planes which contain two long edges
of the pyramid, either opposite or adjacent edges, or both,
whichever give the best-conditioned constructions.
317. Problem. To determine the projections of the
line common to the surfaces of two cylinders, the
axes AB and CD of which intersect at right angles,
CD being vertical.
Horizontal section planes will be convenient.
Describe the semicircle on 00 as diameter, and divide its
circumference into six equal parts.
Consider the section plane whose elevation //;/ passes
through 2. Set off l>2, 1)2 in plan, each equal to m.2 in
elevation, and through the points 2 thus obtained draw
lines parallel to ab. These lines form the plan of the section
of the horizontal cylinder, while that of the vertical cylinder
is the circle, centre c. The two sections intersect in plan
in the four points marked 2, which must therefore be on
the required plan of the line of intersection. Draw projectors
XVI
INTERSECTIONS OK SURFACES
389
from the points in plan to meet bn in 2', 2', which points
are on the elevation of the line of intersection.
Repeat this construction for the planes through the
other points of division of the semicircle and draw curves
through the points so determined, obtaining two curves for
the elevation as shown, while the plan coincides with portions
of the circle, centre c.
Note. If the vertical cylinder had been the smaller one, section
planes parallel to the vertical plane would have been preferable.
39Q PRACTICAL SOLID GEOMETRY chap.
318. Problem. To determine the projections of the
curve of intersection of a given cone and cylinder,
the axes of which intersect each other at right angles,
that of the cone being vertical.
Three Cases are shown. In (i) the cylinder completely
penetrates the cone; in (2) the cylinder and cone circum-
scribe the same sphere ; and in (3) the cone completely
penetrates the cylinder.
Case (1). The projections of the cone and cylinder
are given in the figure, the diameter of the cylinder being
equal to that of a circle described so as to fall within the
elevation of the cone. This circle may be regarded as an
end projection of the cylinder.
Horizontal section planes are convenient, since the
sections of the cone are circles and those of the cylinder
pairs of parallel straight lines in plan. In elevation the
sections are overlapping straight lines.
On g'Ji describe a semicircle and divide its circum-
ference into six equal parts.
Through any of the points of division, say i', draw ////
parallel to xy ; this is the edge elevation of a horizontal
section plane.
With v as centre and radius on describe a circle ; this is
the plan of the section of the cone by the plane LM.
Set off b\ =mi' on each side of b as shown, and draw
lines through the points 1 parallel to ab ; these lines form
the plan of the section of the cylinder by the plane LM.
Hence the points in which these sections intersect are
points on the required plan, their elevations being found by
projectors as shown.
Repeat this construction for the planes through^', i', 2
. . . //, and draw a curve through the points thus found.
The planes through 4, 5', H give rise to points which in
plan are hidden by the cylinder, and the plane through 3'
will divide the dotted and full portions of the curves in
plan. The curve is seen to touch the outline of the cylinder
in plan at these points.
XVI
INTERSECTIONS OF SURFACES
391
Examples on Problem 317.
1. Determine the interpenetration of two cylinders, diameters lY
and 2" ; axes intersecting at right angles, and parallel to vertical
plane ; smaller cylinder horizontal.
2. Work Ex. 1 (a) when the axis of the smaller cylinder is hori-
zontal, but makes 30 with the vertical plane ; (I) when the axis
of the smaller cylinder is inclined at 30 to the horizontal plane.
3. A semi-cylinder, 4V' diameter, rests with its rectangular face
on the ground. A cylinder, 2j" diameter, rests with a generator
on the ground, at right angles to the axis of the semi-cylinder.
Determine the plan of the curve in which the surfaces intersect.
392 PRACTICAL SOLID GEOMETRY chap.
Case (2). The diameter of the cylinder is taken equal
to that of a circle which touches v'd' and v'e'.
Repeat the construction of Case (1). If this be done
accurately, the elevation will be found to be two intersecting
lines. The intersection of the surfaces therefore consists of
two plane sections of either surface, that is, of two ellipses.
These ellipses project as ellipses in plan. In this case the
cone and cylinder circumscribe the same sphere. See
Arts. 228 and 273, and Theorems 23 to 48, Appendix II.
An important section plane is that whose elevation
passes through f, the point where v'e touches the circle;
this section gives the points where the ellipses intersect in
plan and elevation. Observe that at these points the
surfaces touch one another, and two branches of their
intersection cross one another at each point of contact.
Case (3). The diameter of the cylinder is taken equal
to that of a circle described so as to intersect the lines v'e,
v'd', in points such as r, s.
As important section planes take those whose elevations
pass through g and r. Another plane between these will
be sufficient to determine the upper curve of intersection.
For the lower portion select the two important section
planes whose elevations pass through s' and //, and in
addition one or two planes between them.
Observe that the lower curve is hidden by the cylinder
in plan.
Examples. 1. Draw an isosceles triangle abc, with the base
fc=3i", and altitude ad=T, \ this is the elevation of a cone,
vertex A, with its base on the ground. On da take de=i^".
With c as centre describe a circle to touch ah and ac. Also
describe two circles concentric with this, but having radii T V'
less and T V' greater. Each circle is the end elevation of a
cylinder. In each case determine the plan and elevation of
the curve of intersection of the two surfaces, when the cylinder
is turned with its axis parallel to the vertical plane.
2. In Ex. 1 describe a semicircle on xy with centre c and radius
2#" ; this is the end elevation of a semi-cylinder ; draw the
plan of the curve in which the cone and semi-cylinder intersect.
XVI
INTERSECTIONS OF SURFACES
393
I
a'
T?pir l
'\ ry :,Xr' v ^
/ Pi c TNl ' N *
' / ' 1 ' >v^
/ \ / \ v
v *. ''* Nl
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i
1
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V
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y
394 PRACTICAL SOLID GEOMETRY chap.
319. Problem. To determine the projections of the
line of intersection of a given cylinder and cone, the axes
of which are both vertical.
Horizontal section planes are convenient.
Consider a plane whose elevation is tin. With centre
v and radius // describe a circle ; this is the plan of the
section of the cone. The points/,/ in which this circle
intersects the plan of the cylinder are points on the
required plan ; the elevations p', p' are obtained by pro-
jection.
Sections which give important points. Draw vd through
v and a, intersecting the circle, centre a, in d and c ; then
D and C are the highest and lowest points on the line of
intersection. Also draw ni through a parallel to xy ; then
/', // are on the elevation of the outline of the cylinder.
Hence planes should be taken which cut the cone in circles
whose radii are equal to vd, vc, vn, and vi. Two or three
intermediate planes may be required.
Note. Vertical section planes containing the axis of the cone
would be very convenient to take in working this problem ; the
important points D, C, JV, I might be readily determined in this
manner. Such planes would cut both surfaces in straight lines.
320. Problem. To determine the interpenetration of
a given vertical cylinder, axis AA, and a given sphere,
centre S.
Take vertical section planes parallel to XY. Consider
one such plane of which /;;/ is the plan.
Draw the circle with centre s and radius om; this is the
elevation of the section of the sphere. The section of the
cylinder is a pair of straight lines, one of which has c for
its plan and c'c for its elevation. The line and circle
intersect as shown in two points p',p\ which are on the
required elevation, p being the plan of the points P.
Important section planes are those whose plans pass
through e, a, s, and f. One ' or two planes may be taken
in addition to these.
It will be observed that the required plan is the arc erf.
XVI
INTERSECTIONS OF SURFACES
395
Example on Problem 319. Describe two circles with radii i|"
and ", their centres c and v being .85" apart ; these are the
plans of a cylinder and cone standing upright on the ground ;
height of cone 3". Determine the elevation of their curve of
intersection on an xy making 35 with cv.
Obtain the developments of the cone and cylinder, in each
case showing the curve of intersection.
396 PRACTICAL SOLID GEOMETRY chap.
321. Problem. To determine the projections of the
line of intersection of a given sphere, centre S, and a
given vertical triangular prism.
Either vertical or horizontal section planes may be
employed, but the former are preferable in determining the
important points.
Consider the plane whose plan is 1m. With / as centre
describe a circle, the diameter of which is ab ; this is the
elevation of the section of the sphere.
Draw the lines cc, ad' by projecting from c and d ;
these form the elevation of the section of the prism.
The intersections P of these two sections are points on
the required curve.
The important sections. Draw se, s/i, so respectively
perpendicular to the sides of the triangle uzw, and take
section planes through e, o, n ; they give the highest and
lowest points on the three portions of the intersection of
the sphere and prism. Take section planes through u,
s, g } h, s ; the plane through s will give the points on the
outline of the sphere in elevation.
The curves of intersection are ellipses.
Examples on Problems 320 and 321.
1. Describe two circles with radii i-J" and i", the distance between
their centres being |" ; these are the plans of a cylinder and
sphere which intersect. Determine the elevation of the curve
of intersection on an xy which makes 30 with the line joining
the centres of the plans.
2. Draw an equilateral triangle abv 3 V side ; bisect av in c. With
centre c describe a circle to touch ab and av ; this figure is the
plan of a cone, vertex V, and a sphere, the axis of the cone
. being horizontal, and the centre of the sphere on the surface of
the cone. Determine the plan of the curve of intersection
and the elevation on an xy parallel to ab.
3. Draw a triangle abc with ab = 2.7$", ^=3.2", ac = 2.f. Take
a point s inside abc distant 1. 1" from ac and .9" from be ; with
s as centre describe a circle 3.5" diameter. The circle and
triangle are the plans of a sphere and prism. Determine the
elevation of their curve of intersection on an xy making 20
with ab.
XVI
INTERSECTIONS OF SURFACES
397
398 PRACTICAL SOLID GEOMETRY chap.
322. Problem. To determine the plan and an elevation
of the line of intersection of a pyramid and prism, the
indexed plans of which are given ; the pyramid rests with
its base on the horizontal plane, and the edges of the prism
are horizontal. Unit o. i".
Draw an elevation of the solids as seen when looking in
a direction parallel to the long edges of the prism ; that
is, take xy perpendicular to the plans of these horizontal
edges.
In this example the intersection will consist of a series
of straight lines, which may be obtained by determining
(i) the points in which the edges of the prism intersect
the faces of the pyramid; (2) the points in which the
edges of the pyramid intersect the faces of the prism ; and
then joining these points by straight lines in the proper
sequence.
Join v'e and produce to w ; then v'tv is the elevation
of two lines, one on each of the faces A VB, A VD of the
pyramid, if vw, vz are their plans. This may be regarded
as a section by a plane containing J^and the horizontal edge
E. Hence H and K are the points in which the edge E
meets the faces A VB, A VD of the pyramid.
In a similar manner it is found that the edge F
meets the faces CVB, CVD in L and M, and the edge
G meets the same two faces in yVand O.
Next consider the edge VB of the pyramid. It inter-
sects the faces EF and EG of the prism in points whose
elevations are t' and u, the plans t and it being obtained at
once by projection. In like manner the edge VD meets
the same two faces in R and S.
When joining the points thus found, each face of the
prism may be taken in turn and its lines of intersection
with the faces of the pyramid noted.
Thus, taking the face EG, it will be seen that we must
join hit, un, os, sk ; all of which are underneath the prism
and therefore dotted. On the face GEwe get In, mo; and
on the face EE we have ht, //, mr, rk.
XVI
INTERSECTIONS OE SURFACES
399
400 PRACTICAL SOLID GEOMETRY chap.
323. Problem. The indexed plan of an irregular tetra-
hedron is given, the circle representing a cylindrical hole
bored vertically through the solid. Draw the elevation
of the pyramid on a vertical plane parallel to AB.
Take xy parallel to ab and draw the elevation.
The hole passes through the face ABC and partly
through the faces ABD, CBD.
A series of vertical section planes passing through B
will in this case be convenient.
Consider the section plane whose plan is be ; it cuts the
pyramid in a triangle BFE and the surface of the hole in
two vertical lines whose plans are /// and n respectively. As
obtained from the elevations of the lines these two sections
intersect in T, U, M, and IV, which are therefore points on
the required curve.
Important Sections. Through o draw rs parallel to ab.
Then the section planes whose plans pass through /' and
each of the points r, s, d are important ; also the planes
which touch the surface of the hole. In connection with
the last two planes it should be observed that the lines in
which they intersect the faces of the pyramid are tangential
to the curve of intersection. This will appear on drawing
the elevation, and the points of contact should be found.
Examples on Problems 322 and 323.
1. Draw a quadrilateral abed, having a^ = 3f", bc = 2", ac=^",
ad=2", cd=2^". Take an inside point v i" from ab and i-J"
from ad; join v to a, b, c, d. This is the plan of a pyramid
with its base ABCD on the ground, and the vertex V 3" high.
An equilateral triangular prism with its long edges horizontal and
perpendicular to AC penetrates the pyramid. A side of the end
of the prism is 1^" ; one face of the prism is horizontal, its
centre being 2|" vertically below I'. Determine the plan of the
intersection and the elevation on an xy parallel to ac.
2. A quadrilateral a.^b.^c^d^, with the diagonals ac, bd, is the plan
of an irregular pyramid; ab=$", bc ^', ac = 4^", cd=T,",
and ad= 2^". A circle, diameter 2", with its centre on ac and
touching cd, is the plan of a vertical cylindrical hole cut through
the pyramid ; determine the elevation of the pyramid and hole
on an xy parallel to cd. Unit o. 1".
XVI
INTERSECTIONS OE SURFACES
401
2 D
402 PRACTICAL SOLID GEOMETRY chap.
324. Problem. To determine the projections of the
intersection of two surfaces of revolution given in eleva-
tion, the axes of which intersect each other and are
parallel to the vertical plane, one being vertical.
The given surfaces are generated by the revolution of a
circular arc about an axis in the plane of the arc ; in one
case the axis, CD, intersects its arc ; and in the other the
axis, AB, does not intersect it. We shall employ the
method of spherical sections.
Let the axes intersect in O, and with o as centre
describe a circle as shown in the figure ; let this be the
elevation of a spherical section surface which intersects
each given surface of revolution in a pair of circles, each
circle of one pair cutting one of the circles of the other
pair. Thus the circles whose elevations are e'f and g'h'
cut each other at points which have r for elevation.
In like manner s is obtained from the other two circles.
To obtain the plans of these points : with centre o
describe a circle, the diameter of which is g'ti ; draw pro-
jectors from r, s' to meet the circle in r, s.
Repeat this construction for two or three other spherical
sections, centre o.
The plans of the surfaces and of the curves of intersec-
tion are not shown in the figure.
Since the outlines of the surface in elevation may be
regarded as sections by a plane containing the two axes, it
follows that M, N~, U, V are on the curve of intersection.
Examples. 1. Determine the plan and elevation of the curve of
intersection of two surfaces of revolution like those in the figure,
whose axes intersect, the axis of one being vertical and that of
the other inclined at 45 to the ground. The least cross-
section of the one is ii" in diameter, and the greatest cross-
section of the other is i4". The radius of the curved outline of
each is 3^".
2. The axes of a cylinder and a double cone of indefinite length in-
tersect each other at a point ^" from the vertex of the latter.
Vertical angle of cone 55 , axis vertical. Diameter of cylinder
l" ; inclination of axis 45. Draw the plan and elevation of
their curve of intersection.
XVI
INTERSECTIONS OF SURFACES
403
1 a
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\ i '"'' \ >' \//A
Ai7 /r <f r /T\
' / x v / \ A >
K "^"""' "1" c\'\ --
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,-v
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S'h'
404 PRACTICAL SOLID GEOMETRY chap.
325. Miscellaneous Examples.
*1. Fig- () The plans of the edges of a four-faced right prism
with a horizontal axis are given. The prism is penetrated by a
vertical square prism 2|" high. Draw an elevation on a plane
parallel to the axis of the horizontal prism, showing the invisible
portions of the intersection by dotted lines. Unit = o. i".
(1886)
*2. Fig. (b). The elevation of a right cone resting on the horizontal
plane is given. The hatched semicircle is the elevation of a
hemi-cylindrical portion cut out of the cone. Draw the plan
of the remaining portion of the cone. O893)
*3. Fig. (c). The given pyramid on quadrilateral base ABCD in
the horizontal plane, and vertex v. 27 , is penetrated by the given
triangular prism, the edges of which are horizontal. Determine
the intersection of the surfaces of the two solids. Unit = o. 1".
(1890)
*4. Fig. (</). The flat base of a hemisphere of i|" radius (centre O )
rests on the ground, as does also that of a pyramid abed, the
height of its vertex {V) being 2^". Draw the plan and eleva-
tion on xy of the intersection of the pyramid and hemisphere,
and develop the faces BVC, DVC so as to show the develop-
ment of the intersection. ( 1896)
*5. Fig. (e). A vertical triangular slot is cut through a sphere. The
plan of the sphere and slot is given. Draw an elevation of the
sphere. (1879)
*6. Fig. (_/"). It is required to fit a cylindrical steam dome on the
top of a cylindrical boiler shell. A sketch with the required
dimensions is given. Draw the elevation of the curve of inter-
section of the dome and shell, and obtain the development of
the dome. Scale ^ z . (1880)
7. A horizontal cylindrical hole (diameter ij") is bored through a
vertical cylinder (diameter 2|") ; the axis of the boring cylinder
passes j" from the axis of the vertical cylinder and is inclined
2 5 to the vertical plane. Draw the elevation of the vertical
cylinder. ( 1S77)
8. A cylinder and sphere of Ij" and l|" diameters respectively
rest with their curved surfaces on the ground, the centre of
the sphere being in the surface of the cylinder. Draw the
plan of the curve of penetration and an elevation of the curve
on a vertical plane making 45 with the axis of the cylinder.
9. A cone, height 3", rests with its base on the ground, diameter of
base 3". The axis of a cylinder 2" diameter is parallel to the
vertical plane and inclined at 45 to the ground, and passes
through the vertex of the cone. Determine the plan and eleva-
tion of the curve of intersection of the surfaces.
XVI
INTERSECTIONS OE SUKEACES
405
Copy the figures double stye.
8 26 o is
CHAPTER XVII
CAST SHADOWS
326. Preliminary. It is a matter of common observa-
tion that the rays of light which emanate from any source
proceed in straight lines in all directions through space,
except so far as they may be intercepted by opaque objects,
or otherwise influenced. If a surface receive the rays, a
portion of the surface will be deprived of light by the inter-
position of the opaque body. The space devoid of rays
between the body and the surface is the shadow of the
body, and that part of the surface deprived of light is
the shadow cast by the body on the surface, or the cast
shadow. Thus a lunar eclipse occurs when the moon
enters the region of the earth's shadow, and as we watch
the phenomenon we see the earth's cast shadow on the face
of the moon.
The shadows we see around us are generally of a most
complicated nature. Although there may be only one
original source, yet all the surrounding objects on which
the light falls give back some of the light, and thus become
secondary sources of greater or less intensity and of varying
size ; these reflected lights play a very important part in
pictures. We give no account here of this maze of varying
light and shade. Our task is comparatively simple ; we
confine attention to one source of illumination, and this
is further supposed to be so small that it may be treated
as if it were a point. The corresponding shadows may
be termed geometrical. The nearest approach to this in
chap, xvn CAST SHADOWS 407
nature is perhaps an electric arc light, the cast shadows
from which are very sharply defined. When the sun is the
source, the shadows are softened at the edges on account of
the angular magnitude of the sun's disc.
If the point source is near at hand we have divergent
rays, but we generally simplify the problems still further,
and assume that the source is so far away that all the rays
are practically parallel to one another.
327. Theorems relating to geometrical shadows. The
opaque object which casts the shadow receives light from
the source on one portion of its surface, the other portion
being in shade. The line on the surface which divides
the two portions is called the line of separation ; this line
is evidently the locus of the point of contact of those
rays which touch or graze the surface ; it is further evident
that these bounding or extreme rays are those which define
the outline of the shadow cast on any surface. We thus
have
Theorem 1. For any object which casts a shadow, the
line of separation on its surface is the locus of the point of
contact of the bounding rays ; and the shadow cast by the line
of separation on any surface is the outline of the shadow cast
by the object itself on the same surface.
We may if we like regard the rays of light as projectors,
in which case the shadow cast by any body on a plane
becomes its outline projection on the plane. The projec-
tion is parallel (and oblique) ox radial, according to whether
the rays are parallel or divergent.
Under another aspect the cast shadow may be looked
upon as the section of a cylinder or cone, the latter terms
being here used in their wider meanings (see Definitions
24 and 27 of the Appendix); the bounding rays become
the generators of the cylindrical or conical surface, and the
line of separation is the directing curve.
Thus theorems relating to cast shadows may often be at
once deduced from familiar theorems of projection, or of
conic sections, e.g.
408 PRACTICAL SOLID GEOMETRY chap.
Theorem 2. The shadow cast by a point on any surface
is the trace on that surface of the ray through the point.
Theorem 3. The shadow cast by a straight line on any
plane is the straight line joining the shadows cast by its ends.
Theorem 4. The shadow of a straight line on any surface
coincides with the trace on that surface of a plane which con-
tains the straight line and the source, for divergent rays, or
which contains the straight line and is parallel to a ray, for
parallel rays.
Theorem 5. If a series of straight lines be parallel to one
another, their shadoivs on any plane by parallel rays are also
parallel to one another and of proportionate lengths. If the
rays are divergent the shadoivs are also divergent.
Theorem 6. If a plane figure be parallel to a plane, its
shadoiv on the plane is equal and similar to the figure if the
rays be parallel, and is similar if the rays be divergent.
This last theorem follows from a property of the cylinder
or cone, viz. that parallel sections of a cone (or pyramid)
are similar figures, and of a cylinder (or prism) are similar
and equal figures. Thus for parallel rays the shadow cast
on the ground by a horizontal circle is a circle of the same
size, the centre of the latter being the shadow cast by the
centre of the former.
In regard to the general method of working problems,
in some cases it is necessary to determine the line of sepa-
ration before we can obtain the cast shadow ; in others the
shadow helps us to determine the line of separation ; in all
cases the connection between the two should be constantly
borne in mind. Tf the object have any corners, the
shadows of these should be found. When a shadow is cast
on two surfaces which intersect, such as the two planes of
projection, those points on the outline of the shadow which
fall on the intersection should be specially determined as
important points.
The following problems are confined to parallel rays.
The direction of the latter may be defined by the projec-
tions, or by the indexed plan, of any single ray.
XVII
CAST SHADOWS
409
W
<?
(b)
328. Problem. To determine the shadow cast by a
given point P, (a) on the horizontal plane, (b) on the
vertical plane, having given the direction of the parallel
rays in plan and elevation.
(a) Draw the projections of a ray through P ; that is,
through p draw a line parallel to the plan, and through p' a
line parallel to the elevation of the given direction of the
rays. Determine / , the horizontal trace of the ray ; then
f> is the required shadow.
(/?) Draw the projections of a ray through P and deter-
mine its vertical trace p x ; then p x is the required shadow.
Examples. 1. A point A is 2" in front of the vertical plane and
1" above the ground. The plans and elevations of the rays
make angles of 30 and 45 with xy. Determine the shadow
of the point on the ground.
2. Take the point A in Ex. I to be 1" from the vertical plane and
2" above the ground ; find the shadow on the vertical plane.
3. A point A is 2" above the ground and 1" from the vertical plane.
(1) Determine whether the shadow is cast on the vertical or
horizontal plane, the rays being inclined at 40 and their plans
making 30 with xy. (2) The direction of the plan being
unaltered, what must be the inclination of the ray if the shadow
of A is on xy, and what if the shadow is |" below xy?
410 PRACTICAL SOLID GEOMETRY chap.
329. Problem. To determine the shadow cast by a
given line AB, (a) on the horizontal plane, (b) on the
vertical plane, (c) on both planes of projection, having
given the direction of the parallel rays in plan and
elevation.
(a) By Prob. 328 determine a and b , the shadows cast
by A and B on the horizontal plane ; then a b is the re-
quired shadow of the line.
(b) Find by Prob. 328 a l and b v the shadows cast hyA and
B on the vertical plane ; then a x b x is the required shadow.
{c) Determine a b , the shadow cast by AB on the hori-
zontal plane, supposing the vertical plane to be transparent
like glass ; also determine b v B's shadow on the vertical
plane. Let a b meet xy in i ; join t' b r
Then the broken line a / b x is the required shadow, a i
being on the horizontal plane and shown in plan, while i b x
is on the vertical plane and is shown in elevation.
Examples. 1. -/ is a point 2" from each plane of projection ; B
is 3" from the vertical plane and 1" above the ground ; AB is
3" long. Determine (1) the shadow cast on the ground ; (2) the
shadow cast on a plane parallel to the vertical plane and 1" in
front of it ; (3) the shadow cast partly on the vertical plane
and partly on the ground, and determine the point on AB
whose shadow is in xy. The plan and elevation of a ray make
30 and 6o with xy.
2. a-n/'io is 2" long and is the indexed plan of a line AB ; c is 2"
from b and 1" from a. The plan of a ray makes 60 with ab,
while the inclination of a ray is 50 . If the shadow of C falls
on the shadow of AB, determine the index of c. Unit = o. 1".
3. The plan and elevation of a ray both make 45 with xy
determine the inclination of the ray. Am. 35-2.
4. A system of parallel rays make 45 with xy in plan, and their
inclination is 45 ; find their direction in elevation.
5. A triangle ABC has the point A on the ground, and its plan is
an equilateral triangle abc of 2.5" side. The shadow of ABC
on the ground is a right-angled isosceles triangle with a as
vertex, the parallel rays having an inclination of 45, and their
plan making 45 with ab and 15 with ac. Determine the
shadow a^ r ; index the points b and c ; and find the true
shape of ABC.
Hint. Make use of Prob. 21 to draw the shadow.
XVII
CAST SHADOWS
411
Examples on Problems 330 and 331.
1. A cube 2" edge rests with a face on the ground, one side of the
base making 40 with xy, the nearer end of this side being 1"
from the vertical plane. Determine the shadow on the planes
of projections when the rays in plan and elevation make angles
of 45^ with and are directed towards xy. Indicate the two
portions of the line of separation which give rise to the two
parts of the shadow, namely, those which are cast on the
horizontal and vertical planes.
2. Copy double size the figured plan of the tetrahedron on p. 199,
and attach the same indices; unit o. 1". Find the shadow of
the solid on the ground, without drawing an elevation, making
use of the following theorem. The rays are inclined at 45,
and in plan are parallel to ab.
Theorem. For parallel rays, the line joining the plan of a
point to its shadow on the ground is parallel to the plan of a ray,
and of length proportional to the height (or index) of the point,
this length being equal to the height when the inclination of the
ray is 45 .
3. An octahedron of 2" edge rests with a face on the ground.
Draw its plan, and by the method of Ex. 2 determine its
shadow on the ground, the rays being inclined at 35 , and in
plan making 45" with a horizontal edge of the solid.
4. Determine the shadow cast on both planes of projection by the
pyramid in Ex. 23, page 201, the vertical plane being parallel
to ad, the rays inclined at 45 and parallel, in plan, to av
Indicate the line of separation and the portions of it which cast
the shadows on the ground and vertical plane respectively.
412 PRACTICAL SOLID GEOMETRY chap.
330. Problem. A given cube rests with one face on the
ground ; to determine its shadow on the latter from given
parallel rays. Also to find the line of separation on the
cube.
The projections of the cube are shown in the figure.
Consider the vertical edge BB. Obtain b , the shadow of
the upper end B. Join bb ; then bb Q is the shadow of BB.
I )raw cr and dd , each parallel and equal to bb ; join
b c and c d . Then bb t d dab is the outline of the required
shadow of the cube.
It will be evident that the outline of the shadow is cast
from the vertical edge BB, the upper horizontal edges BC,
CD, the vertical edge DD, and the lower horizontal edges
DA, AB ; hence these edges constitute the line of separa-
tion, the vertical faces BC, CD and the base of the cube
being in shade.
331. Problem. The indexed plan of an irregular pyra-
mid, with its base ABCD resting on the ground, is given,
and also r n s 7 , the indexed plan of a parallel ray. To
determine the shadow of the pyramid on the ground, and
on the vertical plane the plan of which is lm. Also to
indicate the line of separation on the solid. Unit 01".
On //// as a ground line determine the elevations of BS
and V. Obtain v Q and v v the horizontal and vertical
traces of a ray through V, and join v Q b, v d, meeting /;;/ in
/o and e .
The lines v b, v d, together with ab, ad, would form the
outline of the shadow of the pyramid on the ground if the
vertical plane LM were transparent or were removed.
Join e v v f f) V 1 ; then b/ e dab is the outline (in plan) of
that part of the shadow which falls on the ground, and
f v x e Q is the outline (in elevation) of the remainder of the
shadow, which falls on the given vertical plane.
Since the outline of the shadow is cast from the edges
VB, BA, AD, and D V, these edges constitute the line of
separation.
XVII
CAST SHADOWS
413
414 PRACTICAL SOLID GEOMETRY chap.
332. Problem. A given circle, centre C, is parallel to
the vertical plane, to determine the shadow cast by it on
the planes of projection, having given the rays.
We must first find the shadow cast on the vertical plane,
on the supposition that the horizontal plane is transparent.
That is, determine c v the shadow cast by C on the vertical
plane, and with c x as centre describe a circular arc with a
radius equal to that of the given circle ; this arc meets xy
in d and e , and is that portion of the required shadow
which falls on the vertical plane. See Theorem 6, Art. 327.
Through d draw d d' parallel to the given elevation of
a ray, and draw d'e parallel to xy. It will be obvious that
the arc DSE is that which casts the shadow on the vertical
plane ; the remaining arc DRE throws its shadow on the
horizontal plane.
To determine the latter shadow, take any point N on
the circle ; determine n ; draw n Q m Q parallel to xy, making
n m = u'm'; then // , m are two points on the required
shadow. Repeat this construction for one or two other
points on arc RJYD, and draw the elliptical arc d ce through
the points so obtained. This will be the required shadow
on the horizontal plane.
Note that at the point of contact of the circle and
ground, the tangents to the shadow and circle coincide.
An important ray is the one which touches the circle in
elevation at say t' (not shown). The corresponding
projector for the shadow touches the latter at / .
333. Problem. To determine the shadow cast by a
given sphere on the ground, the given rays being parallel.
To find also -the line of separation on the sphere.
Suppose the sphere to be circumscribed by a cylinder,
the axis of which is parallel to the given rays, then the
horizontal trace of the cylinder will be the required shadow,
which may therefore be found as in Prob. 277.
The line of separation is the circle of contact of the
cylinder and sphere.
XV II
CAST SHADOWS
415
Examples. 1. A circle 2" in diameter has its plane parallel to
the vertical plane and 1" therefrom, its centre being 1^" above
the ground. Determine the shadow cast on the planes of
projection, the rays in plan and elevation making angles of
35 and 40 respectively with xy.
Determine the projections of that chord of the circle whose
shadow coincides with xy.
2. Suppose that the circle in Ex. i has its plane horizontal, the
centre being 2" above the ground and ii" from the vertical
plane. Determine the shadow cast from it on the planes of
projection.
3. A sphere 2 : ' radius has its centre 2" from each plane of projec-
tion. Determine the shadow on the ground if the rays on
plan and elevation make angles of 30 and 45 with xy.
4. Determine the shadow of the sphere in Ex. 3 cast on a plane
inclined at 30 , and I A" from the centre of the sphere, the rays
being parallel to the vertical plane, and inclined at 40 to the
ground and UO to the inclined plane.
5. In Ex. 4 what should be the inclination of the plane, so that
the plan of the shadow is a circle, the rays being unchanged ?
416 PRACTICAL SOLID GEOMETRY chap.
334. Problem. To determine the shadow cast on the
horizontal plane, (a) by a given cone resting with its base
on the ground ; (b) by a given cone, axis vertical, with
its vertex on the ground ; and (c) by a given cylinder
resting with its base on the ground. Also to show the
line of separation in each case.
(a) Obtain v Q , the shadow of the vertex on the ground ;
draw the tangents v Q f , 7> r Q ; then these tangents together
with the arc / r form the outline of the required shadow.
Draw the radii vt Q , vr Q at right angles to the tangents ;
then the generators VJ\ VR and the arc TRN form the
line of separation on the cone, the portion VTNR V being
illuminated, while the remaining surface, including the base,
is in shade.
(b) Determine c , the shadow of the centre of the base
of the cone, and with c as centre describe a circle with a
radius equal to that of the base ; draw the tangents z// , vr Q .
These tangents together with the arc t^n Q r Q form the outline
of the shadow on the ground.
Draw the radii cJ w c Q r Q perpendicular to the tangents ;
and draw vt, vr respectively parallel to these radii ; then
VT, VR and the arc TNR constitute the line of separa-
tion, and the portion VTNRV of the conical surface is in
shade.
(c) Determine o Q , the shadow of the centre of the upper
end, and with o as centre describe a circle with a radius
equal to that of the cylinder. Draw the tangents tt {) , rr Q ;
then these tangents together with the semicircles rmt, r Q n Q t Q
form the outline of the required shadow.
Draw the diameter rt at right angles to oo Q , then the
generators RR, TT, the upper semicircle RNT, and the
lower semicircle RA/T form the line of separation ; one-half
of the cylinder, including the base, is in shade.
Note. Observe that the shadows z' f , ~'(/o m ( a ) an( l (^), and
rr , tt in (c) are the horizontal traces of the planes parallel to the rays
which touch the surfaces along VR, VT, or RR, TT. See Theorem 4,
Art. 327.
XVII
CAST SHADOWS
417
X (b
yi/'
\c:
V
\
r tif-
,^s
^\77
X
^cn\
' n'n
tor
fWf
lnl \0o v
1 1 j
Examples. 1. A cone rests with its base on the ground; diameter
of base 2", height 3 ", and the centre of the base 2j" from xy.
Determine the shadow on the ground, the rays in plan and
elevation making angles of 30" and 45" with xy. Indicate
the line of separation, and the portion of the surface of the
cone which is in shade.
2. Suppose the cone in Ex. 1 to be cut by a horizontal plane 2"
high ; find the shadow of the frustrum on the ground.
3. Determine the shadow on the ground of the cone in Ex. 1 when
resting with the axis vertical and the vertex on the ground.
Show the line of separation.
4. A cylinder 2" diameter, length of axis 3", rests with one end on
the ground, the centre of the base being 2^" from xy ; if the
rays are as in Ex. 1, determine the shadow on the ground.
Indicate the line of separation.
2 E
4i3 PRACTICAL SOLID GEOMETRY chap.
335. Problem. A given cylinder, axis AC, rests on
the ground with its axis at right angles to the vertical
plane. Determine the projections of its shadow on the
planes of projection from given parallel rays. Show
also the line of separation on the cylinder.
We may regard the required shadow as consisting of
three portions, one from the curved surface of the cylinder,
and two others from the flat circular ends. If these three
were found separately they would be seen to intersect each
other ; only the outline is required for the shadow.
To determine the shadow cast by the curved surface,
draw the two tangential rays in elevation, touching the circle
at /' and r, from which // and rr may be obtained by pro-
jection. Then TT and RR form the line of separation for
the curved part of the cylinder.
Obtain r Q r Q and t e t v the shadows of RR and TT on
the horizontal and vertical planes (Prob. 329) ; in determin-
ing these use may be made of the fact that r r and t t are
each equal and parallel to rr or //.
To obtain the shadow cast by the circular end nearer to
the vertical plane, determine c v the vertical trace of a ray
through C, and with c x as centre and radius equal to that
of the cylinder describe a circular arc ; it will touch e () t 1 at
t x and meet xy in d . This arc is that portion of the shadow
of the circle which falls on the vertical plane.
Through d draw d d' parallel to the elevation of a ray ;
then the arc DR is the only part of the circular end which
casts a shadow on the ground. To obtain this shadow it
will be sufficient to consider one point on DR between D
and R (not shown), then by finding the shadow of this point
we shall be enabled to draw the elliptical arc r d .
The shadow from the other circular end consists of an
elliptical arc extending from r to t Q as shown, and passing
through a. Points on this arc may be found by taking a
few points on the semicircle TNR and determining their
shadows on the ground.
Observe that r (J r t) , t Q f Q are tangents to the elliptic arcs at
XVII
CAST SHADOWS
419
i> '0 >
r a t ft
and that e t 1 touches the circle at t x ; note also
that at and nn Q are tangents to the ellipse at a and n .
The line of separation consists of the generators J?A,
TT, together with the semicircles RNT, RDT. This is
the line whose shadow gives the outline.
Examples. 1. A cylinder, 2" diameter, length 2", rests on the
ground along a generator which is at right angles to the vertical
plane. The nearer end of the cylinder is " from the vertical
plane. The rays, in plan and elevation, make angles of 30'
and 45" with and are directed towards xy. Determine the
shadow on the planes of projection ; indicate the line of separa-
tion, showing the two parts of it which cast the shadow on the
horizontal and vertical planes respectively.
2. A semi-cylinder, 2.5" diameter, 2" long, rests with a semi-
circular end on the ground, its curved surface touching the
vertical plane, and its rectangular face parallel to the latter.
Draw its plan and elevation and determine the shadow cast on
the planes of projection, the parallel rays making 45 with
xy in both plan and elevation, and being directed towards a v.
3. A hemisphere 3" diameter rests with its curved surface on the
ground, its flat face being horizontal, and its centre 2" in front
of the vertical plane. Obtain the shadow cast on the planes
of projection, the rays being as in Ex. 2.
4 20 PRACTICAL SOLID GEOMETRY chap.
336. Problem. A square slab as shown in the figure
has a square hole cut through its centre ; to determine
its shadow on the ground, the parallel rays being given.
Also to indicate the line of separation.
The shadow cast by the outer edges of the slab is easily
obtained and requires no description.
For the hole, the shadows from the upper and lower
squares, supposing that they existed alone, would be
/ O t o and iVyyo respectively. These squares intersect
at two points, denoted by u , v and t , w ; from which we
infer that the two rays which pass through them respectively
must intersect the sides of both the upper and lower squares
of the hole. Draw the plans of the rays through 7' and t
intersecting the plan of the hole in v, u and t, w.
It will now be obvious that the outline of the shadow
from the edges of the hole is cast by TL and L V on the
upper square, and UR, RJV on the lower, and is composed
of the iimermost lines of the squares /<// ?y 0) / //i /i o . On
the contrary the outline of the shadow of the outer edges
of the slab might be obtained by first determining the
shadow from the upper and lower squares and selecting the
outer-most lines for the outline of the shadow.
It should be noticed that the shadow from VM would
be cast on the vertical face MNRQ of the hole and would
be a straight line from U to M ; this is not shown.
Similarly, the shadow from the line TO would be a straight
line WO on the face ONUS.
The line of separation therefore consists of the edges
EF, FB, BC, CB>, DH, HE, and also the lines LM,
MU, UR, RJF, 1 TO, OL. The portions TM, TO cast
shadows on the internal surface of the hole, and there-
fore do not contribute to the outline of the shadow on the
ground.
The student should learn from this instructive example
that the points in which two shadows intersect may assist
in tracing the line of separation, which always passes round
the object in a circuit without break.
XVII
CAST SHADOWS
421
a
I' d' 0' m'b' n!
_i IIIIIMII
:IHllll!lllllll ! llllll||[|llllllillllli;
p' H, s' <f f r
9
\
Example. Copy the above plan and elevation half as large again
as shown, and determine the shadow on the ground when the
rays make 45" with xy in both plan and elevation.
422 PRACTICAL SOLID GEOMETRY chap.
337. Problem. To determine the projections of the
line of separation on the surface of a given solid of
revolution, axis vertical, the rays being parallel to the
vertical plane, and hence to obtain the shadow of the
solid on the ground.
The solid considered is the same as that in Prob. 281.
Let d be the centre for the arc deb'.
Draw any radius o ' d' intersecting db' in s' ; with centre
/ draw the circle through d '. This is the elevation of a
sphere inscribed in the given figure, and touching it in a
circle whose elevation is e'd'. Through / draw a diameter
perpendicular to the given elevation of a ray, and meeting
e'd' in/'. Then P is a point on the required line of separa-
tion.
For the ray through P evidently touches the sphere at P ;
hence it must also touch the surface of revolution at P.
(See Art. 285.) P 'is thus a point on the required line of
separation.
The plan of P may be found by describing a circle with
s as centre, and diameter equal to e'd' ; a projector from />
intersects this circle in />, p, which are the plans of two
points on the line of separation.
Three or four other spheres should be taken, and the
above construction repeated, a curve being drawn through
the points thus found.
The points A 7 " will be on the line of separation. Also,
if a radius of be drawn at right angles to the elevation of
a ray, the highest point F oi the curve will be found, and
this will be on the outline in elevation.
The shadow on the ground is obtained by considering
rays which pass through the points on the line of separa-
tion. For example, the points P give the shadows />, p .
One half of the surface is illuminated.
Examples. 1. Determine the shadow cast on the horizontal
plane by the surface of revolution in Ex. 1, Prob. 282, when
its axis is vertical, one end being on the ground. The rays are
parallel to the vertical plane, and inclined at 45 .
XVII
CAST SHADOWS
423
\ /H.
e'L
p'flm
1. s ',
1/- ' '/ L -t k
/A .'" /
/=- =
o 1 '
C
X
y
Draw a line ab parallel to xy, 1 J" above it and 2^" long. With
a/' as radius, and a and b in turn as centres, describe arcs
commencing at b and a and terminating in xy. This is the
elevation of a surface of revolution ; determine the shadow cast
on the ground, the rays being parallel to the vertical plane, and
inclined at 40.
424 PRACTICAL SOLID GEOMETRY chap.
338. Problem. To determine the shadow cast by the
hexagonal head of the given holt on the cylindrical shank,
from given parallel rays.
In this example the plan of the shadow surface is also
an edge view of the surface, so that the plan of the shadow
is in this edge view.
Draw the tangents ra, tb parallel to the given plan of
the ray ; these represent the extreme rays, and the plan of
the shadow on the shank is the semicircle rpt. It is
evident that the shadow is cast from the portion ACB of
the lower hexagon.
We may choose any point Q (i.e. q, g) in ACB; then
to determine the corresponding shadow P draw gf> the plan
of the ray QP, and project from p to p' on to the elevation
of the ray drawn through q.
We ought now to select the rays which give the im-
portant points. There is one through the angular point C
which gives the angular point S of the shadow.
Another is drawn through n in plan, and this determines
n, a point on the outline in elevation.
Next, to find the highest points in the two elliptic arcs of
the shadow. For the arc PJVS draw on perpendicular to ac
and uv the plan of a ray ; obtain vit' ; then U is the
highest point of the curve RNS, and the tangent at u' is
horizontal. A similar construction would determine P, the
highest point in ST, the tangent at e being horizontal.
The two extreme rays AR and B T illustrate the following
theorem relating to shadows which fall on curved surfaces :
Theorem. A ray which is tangential to the shadow
surface is also tangential to the shadow ; and the plan and
elevation of the ray are respectively tangential to the plan and
elevation of the shadow.
Thus b't', a'r touch the curves at f and r.
Examples on Problems 338 and 339.
1. Copy the figure of Prob. 338 double size, the edge AC being
placed perpendicular to the vertical plane. Then work the
problem, the rays making 45 with xy in plan and elevation.
XVII
CAST SHADOWS
425
C f &
/
1
1
\
\
\
i
1
2. Draw Figs, (a) and (//) to the dimensions given above, and de-
termine the shadows of the heads on the shanks, as in Proh.
33S ; the rays making 45 with xy in plan and elevation.
3. Draw the projections (a) and (/') when the plans are turned
through 45. Then work Ex. 2.
4. The shank of a rivet is a cylinder l" diameter, and the head is
a cone, diameter of base 1-^", height 1". Determine the
shadow of the rivet on the planes of projections, the axis of the
rivet being vertical and 2#" from the vertical plane : the rays
in plan and elevation making angles of 30 and 40 with xy.
Show also the shadow of the head on the shank, and the line of
separation on the solid. Scale, double size.
5. Work examples 1 to 4, supposing in each case the solid shank
to be replaced by a thin hollow shank with the front half cut
away and removed.
426 PRACTICAL SOLID GEOMETRY chap.
339. Problem. To determine the shadow cast by the
head of the given rivet on the shank, and by the rivet on
the planes of projection, the direction of the parallel rays
of light being given. Also to indicate the line of separa-
tion on the rivet.
First, determine the shadow cast by the cylindrical
shank on the ground, supposing- the vertical plane to be
transparent ; it consists of the two tangents kk Q , // and the
semicircles w / , kgl.
Next, obtain the shadow cast by the head on the ground ;
it consists of the circle, centre c Q , and its tangents v r Q , v Q L.
Draw t Q t and r Q r parallel to the plan of a ray ; then VT
and VR form the line of separation on the curved surface
of the cone. The shadow of the cone is therefore that
cast by the generators VR, VT, and the arc RFHT.
Determine v x r v the shadow cast by VR on the vertical
plane ; join zy . Then v x e Q t Q is the shadow cast by VT on
the two planes of projection.
Draw d Q d parallel to the plan of a ray ; then the arc
DR casts its shadow on the vertical plane, and this shadow
may be found by taking one or two points in DR and
obtaining the shadows cast by them ; the construction is
shown for one such point N, which gives the shadow n v
The outline of the shadow on the horizontal plane is
kutfltfjtfvjgk, and on the vertical plane is d^n^r^v^e^d^.
To obtain the shadow of the head on the shank, draw
the tangents Ih and //" parallel to the plan of a ray; these
are the limiting rays, and the arc FH is that which casts
the required shadow.
Take any point q on fh and draw qp, the plan of a ray
through Q. This ray is intercepted by the surface of the
cylinder at a point on the generator whose plan is p ; draw
the elevation of this generator, and through q draw q'p'
parallel to the elevation of a ray and meeting the elevation
of the generator at /', then p' is on the elevation of the
shadow.
Repeat this construction for a few other points on fh,
XVII
CAST SHADOWS
427
not forgetting to determine the important points /", s', g
and the highest point on the curve. (See the remarks
towards the end of the last problem.) The elevation of the
shadow on the shank is the curve s'p'g'i'.
The line of separation consists of two detached portions,
each closed, viz. LSGIKL and VTHRV.
428 PRACTICAL SOLID GEOMETRY chap.
340. Problem. Having given any two lines AB and
CD, and the direction of the rays of light ; to find the
two points F and E in which a ray intersects both lines ;
that is, to find the shadow of the one line on the other.
Method. Find the shadows of both lines on any con-
venient surface. Project the ray through the point where
the shadows intersect. This ray will cut the given lines in
the required points.
In the figure the shadows aj> w c d on the horizontal
plane are found ; these intersect in the point marked e / ;
a ray through this point is seen to cut the given lines in the
points Eand E. Then the point E is the shadow of AB
on CD, and F is the point in AB which casts the shadow.
Note.- The student should make careful note of the principle
underlying this problem, and of the corresponding construction. The
lines AB and CD may be curved. AB typifies the line of separation,
and CD a selected line on the shadow surface ; the construction shows
how to find the point where the shadow cuts CD. See applica-
tions in following problems.
341, Problem. Having given a line AB, and a thin
circular plate, centre C, parallel to the ground, to determine
their shadows on the ground. Also to obtain the shadow
of the line on the plate, and the portions of the line which
cast the shadows on the plate and ground.
Obtain aj> , the shadow of the line on the ground, on
the supposition that the plate is removed; find also the
circular shadow, centre c , cast by the plate on the ground.
These shadows intersect each other in e , f , and g Q , // .
Thus the outline of the required shadow on the ground
consists of the circle, centre c , and the lines a / and /> /i .
Through e and g draw the lines e Q ef and g g/i parallel
to the plan of a ray ; then, as in Prob. 340, the rays which
meet the ground in e and g intersect the circular plate in
E and G, and the line AB in F and H. Hence EG is
the required shadow on the plate, cast by EH; and AE
and BH throw their shadows on the ground.
XVII
CAST SHADOWS
429
As'
\
X
fj
y
iv?
'
e'
'! ,r .
!:! 9,
\ ~^-
341
43o PRACTICAL SOLID GEOMETRY chap.
342. Problem. The plans of a line EF and of a square
pyramid VABCD are given, the points v, e, f being
indexed, and the base ABCD resting on the ground.
Determine the plan of the shadow cast by the line EF on
the ground and on the pyramid, rs being the plan of a
ray, the inclination of the rays being a.
Draw xy parallel to rs, and project the elevations rs,
e'f\ and*/. Obtain e Q / , the shadow of FF on the ground,
supposing the pyramid to be removed ; also determine r a,
the shadow of VA on the ground. These two shadows
intersect in ///, and the corresponding point M on EF casts
the shadow P on VA.
The shadow of F will obviously fall on the face VAD,
and to determine its projection join v f and produce to
meet ad in o ; join ov, meeting f f in q. Then q is
evidently the plan of the shadow of F oxs. the pyramid.
The complete shadow of EF is ? u flq in plan.
Examples on Problems 340 to 343.
1. The x, y, z co-ordinates of two points A and B are (i", i", i")
and (3", 3", 2"), and those of C and D are (2", 1", 1") and (1 ",
3", 2"). If "the plan and front elevation of a ray make angles
of 30 and 45 with the axis of Y, determine the projections and
co-ordinates of the points E and F in which a ray intersects
both of the lines AB and CD.
2. Determine the projections of the line parallel to the vertical
plane of ZX, which intersects the lines AB and CD in Ex. 1,
and is inclined at 6o to the ground plane.
3. The two ends A and B of a line 3" long are respectively 2" and
1" from each of the two planes of projection. A horizontal
circular plate, if" diameter, has its centre 1" vertically beneath
the mid-point of AB. Determine the portion of the line AB
which casts a shadow on the plate, the plan and elevation of a
ray making 45 and 6o respectively with xy.
4. An equilateral triangular plate, ih" side, is in a horizontal
position 2" above the ground, the nearest corner A being i-|"
from the vertical plane, and one side making 70" with the
plane. A square plate, 2" side, is also in a horizontal position
with one side at right angles to the vertical plane, one end of
this side being 1" therefrom, 1" to the right of A, and 1"
below it. Determine the shadow of the two plates on the
XVII
CAST SHADOWS
431
m/~...
\ ^ ^-- //
"s.
/ \
K
ground, and the shadow of the one on the other, the rays
making 30 and 45 with xy in plan and elevation.
5. A square pyramid, side of base 2", height 2^ ", has its base on
the ground, one side making 30 with xy, the nearest corner of
the base being 1" from xy. The end E of a line EF is 3V
to the left of the vertex, J" from the vertical plane, and 3"
above the ground ; the end F is 2" to the left of the vertex, 3''
from the vertical plane, and 2" above the ground. Determine
the plan and elevation of the shadow of EF on the ground and
pyramid, the rays being parallel to the vertical plane and in-
clined at 40 to the ground.
6. Describe two circles with a ar.d b as centres and radii " and 1",
ab being 2". The circle with centre a is the plan of a circular
plate 3" above the ground, the other circle being the plan
of a sphere whose centre B is 1^" above the ground. Deter-
mine the portion of the edge of the plate whose shadow falls on
the sphere, the rays being inclined at 50 , their plans making
io 3 with ab.
PRACTICAL SOLID GEOMETRY CHAI'.
343. Problem. The projections of a straight line RT,
and a sphere, centre S, are given ; to determine the pro-
jections of the portions of the line which cast their
shadows on the sphere and ground respectively, the given
rays being parallel to the vertical plane.
The rays which touch the sphere will generate a cylinder
the elevation of which is obtained by drawing the tangents
g'l and tid parallel to the elevation of a ray ; this cylinder
touches the sphere in a circle whose elevation is the diameter
g'h', and this circle is the line of separation en the sphere.
Make Iu=g'ti ; draw um perpendicular to xy and join
////. Regard Im as the edge elevation of a plane ; this
plane will cut the cylinder in an ellipse whose major axis
LM is parallel to the vertical plane, the minor axis being-
equal to the diameter of the cylinder. Now since it has
been arranged that the plan of the major axis is equal to
the diameter of the cylinder, the plan of the ellipse will be
a circle the centre s 2 of which is projected from s 2 , the
middle point of /;//.
The shadow of the sphere, that is of its line of separa-
tion, on the plane LM is the ellipse referred to ; hence
the plan of the shadow is the circle, centre s.,.
Next, obtain the plan of the shadow of the line RT on
the plane LM. That is, draw r'r 2 and /?./ parallel to the
elevation of a ray ; draw the projectors r 2 r 2 and t 2 \ to
meet rr 2 and tt 2 in r 2 and t 2 respectively, then r/ 2 is the
plan of the shadow of R T on the plane LM.
The two shadows intersect in e 2 and/,. Now since e 2 is
on the circle, centre s. the ray through E 2 touches the
sphere ; also since e., is on r 2 t 2 the ray through E. 2 intersects
RT. Similar remarks apply to /,. Hence the rays which
pass through E 2 and F 2 respectively touch the sphere and
intersect RT. "The projections of these rays are e 2 e 3 e,
e^e' and / 2 / 3 / f^f, F 3 and F 3 being the points of
contact with the sphere.
Therefore the portion FFcasts its shadow on the sphere,
and FT, FR cast their shadows on the ground.
XVII
CAST SHADOWS
433
Example. Describe a circle, centre c, radius i", and draw a line
add to touch the circle at d ; make ad = \" , and db = \".
Attach indices of 30, 20, and 15 to a, b, and c. Find the
part of the line AB whose shadow falls on the sphere C, the
rays being inclined at 45 and making 6o with ba in plan.
Determine also the shadow cast on the sphere. Unit o. 1".
2 F
434 PRACTICAL SOLID GEOMETRY chap.
344. Problem. The projections of a truncated cone
with a circular slab placed centrally upon it are given ;
it is required to determine the projections of the shadow-
cast on the cone. The parallel rays are given.
The shadow is cast from the lower edge of the slab, but
the exact portion of this edge which casts the shadow on
the cone is not so readily found as was the case with EQH
in Prob. 339. The method now adopted is of general
application, and might have been used in Probs. 338 and
339, had it been necessary to do so.
Obtain v the elevation of the vertex of the cone suppos-
ing the truncated portion to be extended ; and by Prob. 334
determine the projections of the two generators VT, VR,
which form the line of separation on the cone ; this is done by
first obtaining the shadow of the cone on the ground. Now
draw the shadow cast from the lower circular edge of the
slab ; it is the circle with centre c . This circle intersects
the shadow of the cone in the points marked e / and g h Q .
Draw e /e and g Q hg, the plans of the rays which meet
the ground in e Q and g respectively ; then, by the principles
explained in Prob. 340, these rays intersect the generators
VR, VT in jFand H, and the lower edge of the slab in E
and 67 respectively. Consequently the arc EG is that which
casts the required shadow on the surface of the cone.
Draw ov, o'v the projections of any generator O V, and
draw ov the plan of the shadow of O V; let p g be the
point where oz> intersects the shadow of the arc EG.
Draw pQpq parallel to the plan of a ray, then the ray
through p intersects OV in P and the lower edge of the
slab in Q ; and therefore P is a point on the outline of the
required shadow. The elevation /' may be found by pro-
jecting from/ on ti o'v at /', or, as this is not very well
conditioned, by projecting p to p Q ', then drawing p 'p' the
elevation of the ray through p '.
This construction should be repeated for other genera-
tors similar to OV. The generator A V will give n on the
outline in elevation. The two extreme rays EE, GH touch
XVII
\
CAST SHADOWS
V'
1 \\
the surface of the cone, and their projections are tangential to
the projections of the shadow at / and g. The projec-
tions of N, F, and H should be obtained as important points.
436 PRACTICAL SOLID GEOMETRY chap.
The preceding construction may be slightly varied in the
following manner. Choose q, draw qp^ v Q p o, ov, the latter
meeting qp in p. Draw p Q p ' and p Q 'p' to meet a projector
from / in p' ; in this manner, however, JV would not be
determined unless it were found accidentally.
The elevation of the outline of the shadow on the cone
is the curve f'n 'p'h', and the plan is the curve fnph, while
HT and FR are portions of the line of separation on the
conical- surface.
Instead of taking generators such as O V, circles may
be selected on the surface of the cone, in which case their
shadows must be drawn and the points obtained in which
these intersect the shadow of the lower edge of the slab.
By this method, however, the limiting rays EF and GH
could not be accurately determined.
345. Problem. To determine the shadow cast by a
given point P on a given oblique plane VTH, from given
parallel rays RS.
We must find the point where the ray through P meets
the given plane. This may be done as in Prob. 199, or by
either of the following methods.
(a) Through p' draw p'tm parallel to r's', and draw mn
perpendicular to xy. Then I' mn are the traces of an in-
clined plane which contains the ray through P.
Obtain nl, the plan of the intersection of the planes
VTH, LMN ; and draw pq parallel to rs to meet In in q.
Project from q to q .
Then Q is the intersection of the ray PQ and the plane
VTH, and is the required shadow on the plane.
(b) Draw ////;/ parallel to rs, and nil' perpendicular to
xy. Project from n to n ; join I'ri . Draw p'q parallel to
r's ; and project from q to q.
Then Q is the required shadow of P on the plane
VTH.
In this case a vertical plane containing the ray PQ
has been substituted for the inclined plane of (a).
XVII
CAST SHADOWS
437
Note. The problem of finding the intersection ofa given lineand plane
is generally best solved by applying the construction of either (a) or (b).
Examples on Problems 344 and 345.
1. A cone rests with its base on the ground ; diameter of base
2\", height 5". It is cut by a horizontal plane which bisects
the axis, and the upper portion is removed. ' A circular slab, 3"
diameter and f" thick, rests centrally on the frustum of the
cone. Determine the plan and elevation of the shadow cast
from the slab on to the conical surface, the rays in plan and
elevation making 30" and 50 with xy.
2. The traces vt, th of an oblique plane make angles of 6o and
40' with xy. A point P, in a plane through / at right angles
to xy, is 2" from each plane of projection. Determine the
shadow of P on the oblique plane, if the rays in plan and
elevation make angles of 30^ and 45 with xy.
3. Taking the plane VTH and the point /'as in Ex. 2, let PQR
be an equilateral triangular plate, iV edge, in a horizontal
position with PQ parallel to xy and directed towards the
plane. Determine the shadow of the plate cast on the ground
and on the plane, the direction of the rays being unaltered.
4. An equilateral triangular plate ABC, 2" edge, has one edge
AB on the ground at right angles to the vertical plane, the
nearer end of the edge being 1" from xy ; the plate makes 40"
with the ground. A line EF has its end E 1 .V to the left of
AB, 2" above it, and 2" from the vertical plane. The end F
is -i" to the left of AB, 1" above it, and 3" from the vertical
plane. If the rays in plan and elevation make angles of 30
and 45" with xy, determine the shadow of the line EF on
the ground and on the triangular plate.
438 PRACTICAL SOLID GEOMETRY chap.
346. Miscellaneous Examples.
1. A tetrahedron of iV' edge has its three lowest corners o. i", i",
and 1.3" respectively above the horizontal plane. Show the
shadow thrown by it upon that plane, assuming the rays of
light to be inclined at 45 , and their plans to make an angle of
45 with the horizontal trace of the plane containing the three
given points. (1881)
*2. Fig. (a). The plan of an octagonal prism, .V thick, with a
square hole cut through it, is given. Assuming the height of the
lower surface of the prism above the horizontal plane to be 1-i",
obtain the complete outline of the shadow thrown on the hori-
zontal plane. N.B. The rays of light are parallel, and their
direction is given in plan and elevation. 0893)
*3. Fig. (/;). The elevation and plan of a chimney are given ; ab is
the plan of one of the parallel rays of light. Determine in plan
the shadow cast upon the roof. Unit = 0.1". (1891)
*4. Fig. (<). Draw the projections of the given cylinder and trun-
cated cone four times the size of the diagram, and determine the
outline of the combined shadow of the two surfaces, thrown by
parallel rays, the direction of which is given, on the horizontal
plane. The shadow of the cylinder on the cone is not required.
(1890)
*5. Fig. (</). The plan is given of a solid prism 1^" deep resting
on the horizontal plane, r is the plan and r the elevation of
a ray of light, to which the other rays are parallel. Show
the shadow thrown on the horizontal and vertical planes.
(1882)
*6. Fig. (e) shows the plan and elevation of an inclined square plate
out of which is cut a rectangular piece. Determine the por-
tions of the shadow cast on the planes of projection. The plan
and elevation of one of the parallel rays are shown.
7. Determine the shadow cast on the horizontal plane from the cap-
stan shown in Fig. (_/), p. 485. Take the rays parallel to the
vertical plane and inclined at 35 to the ground ; also take the
axis of the capstan 2" from the vertical plane. Show the line
of separation on the surface.
*8. Fig. (c). Copy the figure four times the size shown. Deter-
mine the shadow cf the cylinder on the cone if the rays are
parallel to the vertical plane and slope downwards to the right
at 30 . Also obtain the shadow of the cone on the cylinder,
if the rays are parallel to xy and directed to the left.
9. A hollow cylinder 2" long, 3" and 2t>" external and internal
diameters, is cut into two halves by an axial plane. One of the
portions is placed with its end on the ground and the middle
generator of its convex surface against the vertical plane.
XVII
CAST SHADOWS
439
Copy the figures doubl& siz&
44Q PRACTICAL SOLID GEOMETRY chap.
Determine the shadows on the concave surface, and on the
planes of projection, the parallel rays making 45" with xy in
plan and elevation.
*10. Fig. (a). The figure represents a fluted column with a circular
cap. Draw on the elevation the shadow thrown on the column
by the cap. The direction of the parallel rays of light is
given by arrows. ( 1 S97)
""11. Fig. (p). The diagram, represents a horizontal square block,
ABCD, intersecting a square-based pyramid VEFGH. (The
overlapping portions of the solids are left dotted on the dia-
gram.) Determine the intersections of the solids in plan and
elevation ; and draw the outlines of the shadows thrown by
them, one on the other, and by both on the horizontal plane.
The arrows indicate the direction of the parallel rays. (1895)
" ;t "12. Fig. (V). abc is the plan of the base of an octahedron lying in a
horizontal plane iV above the ground ; fcjms that of a pyra-
mid resting on the horizontal plane, its vertex (0) being 2.35"
above the ground. The direction of parallel rays of light is
shown in rjlan by r, their inclination to the horizontal plane
being 44 .
Show in plan the shadow of the octahedron cast on the
ground and on the pyramid. Shade lightly all portions of the
latter not illuminated. (1S96)
*13. Fig. ((/). The given figure is the elevation of a square-headed
bolt with a conical shank standing on the horizontal plane.
The projections r, ;' of one of the parallel rays of light arc-
given. Draw the bolt full size, and obtain the shadow cast on
the horizontal plane and on the shank. (1887)
*14. Fig. (e). Draw the plan, and determine the shadow cast on
the ground by the trestle of which an end elevation is shown.
The length of the top block is 4' 6", and it projects 6" beyond
the legs at each end ; the legs are square in section. Let the
plan and elevation of one of the parallel rays of light be parallel
to each other and make 45 with xy. Scale -jV.
r 15. Fig. (/"). A spherical segment rests on the top of a truncated
hexagonal pyramid. The elevation and plan of the base are
given. Draw the figure four times the given size, and determine
the portion of the solid in shadow, as seen in elevation. The
direction of the parallel rays of light is indicated. (1S88)
16. A right circular cone, 3" high, rests with its base (3" diameter)
on the ground. A sphere i-t" diameter has the vertex of the
cone as its centre. Determine the shadow cast by the sphere
on the cone, the rays being parallel to the vertical plane and in-
clined at 45 to the ground,
XV11
CAST SHADOWS
4M
Copy the figures double size
f
<?'
J-' /
V s
b'ci' \/!\\ c'd'
I/aTTl
I I \ \
c'd'
e' f it g' .
-Z'9-
CHAPTER XVIII
METRIC PROJECTION
347. General explanation. We have seen how by an
indexed plan the form or position of an object may be
defined by one projection only. In this chapter we develop
another method of representation by means of a single view.
And in this case, as in the former one, the projection of
the object is one that can be readily scaled for the purpose
of ascertaining the dimensions of the parts. Each system
is associated with a class of examples coming within its
special province, and with which it is well adapted to deal.
Thus the shape of an irregularly curved surface is well
exhibited by the method of figured plans. AVe have a
well-known example in maps with indexed contours, which
indicate the configuration of the hills and valleys. A similar
series of sections or contours is employed by naval architects
in representing the shape of the surface of a vessel.
Metric projection is well fitted to show the forms and
dimensions of what may be termed j-ectatigular solids, such
as are many examples of wood work ; they are bounded
mainly by three systems of planes mutually perpendicular,
intersecting in three systems of parallel lines in directions
also mutually at right angles. The metric projection of
such an object resembles a perspective view, and like the
latter conveys a realistic impression of the form even to
the uninitiated. The metric view may look distorted ;
but there is compensation in that it is a scale drawing.
chap, xvm METRIC PROJECTION 443
348. Metric scales and axes. Let oa be the projection
of a line OA on any plane. On OA set off any scale, say
a scale of inches, and project the scale on oa.
The latter scale may be used citlier (1) to ascertain the
length of any line parallel to OA by measuring the length
of its projection on the plane, or (2) to set off the length
of the projection, knowing the length of the line.
Such a scale is called a metric scale ; the plane of pro-
jection is called the metric plane ; the direction oa in the
plane is a metric direction ; any line in the plane and par-
allel to oa is a metric line ; and any line of reference
parallel to oa and in the plane is a metric axis.
Suppose an object like a building brick is projected on
the metric plane. The edges of the solid form three
systems of parallel lines ; as each system will have its own
scale, we shall require three scales in measuring all the lines
of the projection. These are known as the trimetric scales.
Let any three axes of reference OX, O Y, OZ be taken in
space parallel to the three systems of lines, then their pro-
jections ox, oy, oz on the metric plane are called metric
axes of the projection. As was explained in Chapter VI L,
the axes OX, OY, OZ serve to define the position of a point
in space. Thus if the co-ordinates of a point P in space
are X, Y, Z, the point may be reached from O by going
first a distance X along OX, then a distance Y parallel to
OY, and finally a distance Z parallel to OZ. And it is
easily seen that the projection p of the point P may be
plotted on the metric plane by stepping off first from a
distance x along ox, then at the end of this a distance y
parallel to oy, and finally a distance z parallel to oz, where
x,y, z are the co-ordinates X, Y, Z, measured on the metric
scales.
Examples. 1. If a line 2.31" long he inclined at $S.f to the
metric plane, find the length of its metric projection.
Ans. 1. 81".
2. Construct a half-size metric scale for a system of parallel lines
inclined at 55" to the metric plane. The scale is to read
inches and eighths of an inch.
444 PRACTICAL SOLID GEOMETRY chap.
349. Isometric projection the axes and scale. Sup-
pose that in a trimetric system the three directions of the
lines in space are all equally inclined to the metric plane,
then the three scales become identical, and only one scale
is required in measuring all the lines of the system in the
metric projection.
In this case we have an isometric or equal-scale system;
the projection on the metric plane is called an isometric pro-
jection, and the scale used is the isometric scale.
The simplest example of such a solid is a cube; and the
edges of a cube will evidently be all equally inclined when
a diagonal of the solid, that is, a line joining two opposite
corners, is perpendicular to the metric plane. This pro-
jection of a cube may be readily determined by ordinary
Descriptive Geometry, but is still more easily obtained by
the method of isometric projection as follows.
To project a cube isometrically. From considerations of
symmetry we see that the three edges of the cube which
radiate from the corner O farthest from the metric plane
will project into three lines radiating from a point o at
equal angles of 120 ; these are very conveniently drawn
with the 30 and 90 angles of the set-square, as shown at
ox, oy, oz in the figure. These lines define the three iso-
metric directions ; and they, or any other concurrent parallel
set, may be taken as isometric axes.
Next, since the edges OA, OB, OC of the cube are equal
and equally inclined, their projections are equal in length.
Therefore mark off along ox, oy, oz three equal lengths
oa, ob, oc. This length is at present undefined.
Finally, to complete the isometric projection of the cube,
we make use of the principle that parallel lines have parallel
projections. Thus to complete the face of which oa, ob
are sides, draw af and //parallel respectively to ob and oa,
and intersecting in f. The corners d, e, g, with the sides
radiating from them, are similarly determined.
The outline of the projection is seen to be a regular
hexagon.
XVIII
METRIC PROTECTION
445
W e must now find the length of the edge of the cube,
of which the figure just obtained is the isometric projection.
The corners A, B, C of the cube are evidently equi-
distant from the plane of projection, for they are the ends of
lines which all start from a common point O, and are equal
and equally inclined. Thus the triangle ABC is parallel to
the metric plane, and abc gives its true shape. Thus any
side of the latter, say ab } is the true length of a diagonal of
one face of the cube.
44* PRACTICAL SOLID GEOMETRY chap.
Draw aF, bF at 45 to ab ; then aF is a side of the
square of which ab is a diagonal ; so aF is the required
length of the edge of the cube. The following method of
constructing an isometric scale is thus suggested :
To construct an isometric scale. From any point a 1 draw
a l F 1 parallel to aF, that is, with the 45 set-square; also
draw rtj/j parallel to of, that is, with the 30 set-square.
On a 1 F 1 set off a true length scale, and project the scale on
a.f, by lines parallel to Ff, that is, lines drawn with the
90 set-square. Then the projected scale a 1 / 1 is the re-
quired isometric scale. Measuring oa on this scale, the edge
of the cube is seen to be 2.95".
To project a rectangular prism. Employing this scale,
we have at the lower part of the figure, drawn half size,
two isometric projections of an ordinary building brick
9" x 4J" x 3". The process is quite simple. From any
point draw three isometric lines ; set off along them from
the point the length, breadth, and thickness of the solid,
using the isometric scale ; complete the figure by drawing
the system of parallel lines, as explained for the cube.
To express the isometric scale numerically. Since the angles Fall,
fall are 45 and 30, we have
Fa sji fa 2
I ~ ' and ^7 = "7=
ah 1 an N /o
af 2 \'2 sj~2 9
hence -4^= -s == = .817 = nearly.
af x / 3 1 v / 3 11
This gives the ratio of the isometric length to the true length.
We are not compelled to employ an isometric scale in
setting out an isometric projection ; an ordinary scale may
be used. But in this case the projection will be that of an
object larger in the inverse ratio of the isometric scale.
However, if this be recognised, and the ordinary scale used
in measuring the projection, no error or ambiguity will
result, and labour will be saved by using an ordinary scale.
win METRIC PROJECTION 447
350. Examples. Represent in isometric projection the following
ten objects :
1. A cube of 3" edge.
2. A building brick 9" x 4!" x 3". Scale .1.
3. A 24" square slab, 3" thick. Scale J.
4. A 1" square prism, 3" long.
5. A box without lid, 6" long, 4" wide, 2" deep inside, and -J" thick
throughout. Scale -\.
6. An instrument box with the lid open at right angles. Dimen-
sions of box outside : length 6", breadth 4V, depth of box ij",
depth of lid J". Thickness of wood at sides and ends ^"; at
top and bottom \". Scale -Jr.
7. The slab of Ex. 3, when pierced with 6" square hole through
its centre.
8. The chimney shown in Fig. (b), page 439.
9. The trestle, Fig. (<), page 441. Scale -J.
10. The column with cap, Fig. (a), page 441.
11. Represent in isometric projection the three co-ordinate planes of
Fig. 160, page 179, and in this view plot
(a) The point A whose co-ordinates are (2", ii", 1").
(b) The point B whose coordinates are (2. 1.5", 1").
(c) The line CD joining the points (1", 2", 3"), (3", 2.5", 2").
(if) The line is/'' joining the ooints (1", - 2", 3"), ( 3",
2-5", -2").
(e) The irregular tetrahedron whose angular points G, H, K, L
are (2", iV, 1"), (1", 2", 3"), (1", 2.5", 1.5"), (1.5",
5", 3- 5")-
(/) The traces of the plane whose intercepts OA, OB, OC
are 2", 3", and 4" respectively.
12. A room is 24 feet long in the direction north and south, 18
feet wide east and west, and 12 feet high. The following are
points within the room :
(a) A, situated 4 feet above the floor, 7 feet from the north
Avail, and 5 feet from the west wall.
(/') B, 3 feet high, 6 feet and 8 feet from the south and
east walls.
(<-) C, 4 feet below the ceiling, and 1 5 and 9 feet from
the south and west walls.
(d) />, 5 feet below the ceiling, and 3 and 12 feet from
the north and east walls.
Draw the room in isometric projection, and in this view
plot the projections a, b, c, if of the points A, B, C, D.
Nole. An isometric scale should be constructed and used in some
of the above examples. Afterwards an ordinary scale may be
employed to set off isometric dimensions.
448 PRACTICAL SOLID GEOMETRY chap.
351. Problem. Required the isometric projection of
a cube of given edge which has a sphere of given radius
in contact with the centre of one face, and a circle in-
scribed in the same face ; a cone of given axis with its
base inscribed in a second face ; and a cylinder of given
length with its base inscribed in a third face.
The cube. Let OA be the given edge. Determine On
the isometric length of the edge. Set out the projection
a, a, ... of the cube, as explained in the last article.
For convenience refer to the visible faces as the upper, and the
right and left front vertical faces. "We shall require the centres c ; the
isometric bisectors dd ; and the diagonals aa of these faces.
The sphere. Let the sphere rest on the centre of the
top face, and let OB be its given radius. Obtain the
isometric radius Ob, and set this up vertically from c to b.
With b as centre, radius OB, describe a circle. This circle
is the required isometric projection of the sphere.
Note I. The diameters of the sphere and its projection are equal.
The circle. Draw a circle DD of the given diameter
OA, and circumscribe this by a square AA ; draw the
diagonals AA, AA intersecting the circle in Jlf, M, N, N.
Draw the equal perpendiculars MK, NK and obtain their
isometric length Ok. Mark off this length at ak, ak, . . .
from the corners of the upper face along the sides as
shown, and through the points k, k, . . . draw the iso-
metric lines intersecting in ///, ///, >i, n. The ellipse drawn
through the eight points ;//, . . n, . . d . . is the required
projection of the circle inscribed in the face. The lines
dd may be called the isometric diameters of the circle.
Note 2. Observe that this is an application of co-ordinates. The
eight points M . . N . . D . . in the circle are referred to the sides
of a circumscribing square A A. The co-ordinates are reduced by the
isometric scale, and then plotted on the projection a, a of this square.
Any plane curve can be similarly referred and plotted. And any ir-
regular solid figure can be referred to a circumscribing rectangular prism,
and its points then plotted on the isometric projection of the prism.
The cone. Let O V be the given length of axis of the
cone, and let the base be inscribed in the left front vertical
face. Draw the ellipse inscribed in the projection of the
XVIII
METRIC PROJECTION
449
K A
2 G
45o PRACTICAL SOLID GEOMETRY chap.
face in the manner just explained. Draw cv equal to Ov,
and in the proper isometric direction, to represent the axis
CV oi the cone. Draw the tangents to the ellipse from
v, the projection of the vertex.
The cylinder. Let the cylinder have its base inscribed
in the right front vertical face of the cube, and let OE be
the given length of the cylinder. Inscribe the ellipse in
the projection of this face ; this is the projection of one
end of the cylinder. To obtain that for the other end,
conceive the ellipse to be moved parallel to itself in the
isometric direction proper to the axis CE of the cylinder,
through a distance Oe equal to isometric length of the axis.
The construction suggested is to draw isometric lines
through a number of points on the first ellipse, marking off
on them lengths each equal to Oe ; we thus obtain points
on the second ellipse. Some of these lines are shown at
// in the figure. The projection of the cylinder is com-
pleted by drawing the common tangents to the two ellipses.
Note 3. Second /net hod of projecting a circle.
Required the isometric projection of a circle, centre c',
radius OE, the plane of which is parallel to the right front
vertical face of the standard cube.
All the lines dd, mm, nn on any face of the cube may be drawn
with one or other of the 30 , 6o, and 90 edges of the set-square.
First, through c draw the isometric lines dd, dd (whose directions
are readily seen by reference to a standard cube) ; and along them set
off the four isometric radii e'd, each equal to Oe. Next draw the lines
through c in the directions of the major and minor axes. On the
former (that bisecting the acute angles between dd, dd) set off cm, cm
each equal to the true radius OE. Finally through m, m draw the
isometric lines intersecting in 11, n.
Then mm, nn, are the major and minor axes, and vinmn is an in-
scribed square. We thus obtain the projection by drawing the iso-
metric diameter, the major axis, and the inscribed square. Before
sketching the ellipse through the eight points thus found, it is well to
draw short lengths of the tangents at these points as shown.
Note 4. If an ordinary scale be used to set off isometric dimensions,
the radius of the circle which represents a sphere, or the semi-major
axis of the ellipse which represents a circle, are larger than the actual
radius in the inverse ratio of the isometric scale.
xviii METRIC PROJECTION 451
352, Examples. Represent by an isometric projection :
1. A sphere of 1 \" diameter.
2. A circle 2" diameter.
3- A cone, diameter of base 2.2", length of axis 1.7 ".
4- A cylinder 2.5" diameter, .6" long.
5. A grindstone, 24" diameter, 5" thick, with a 6' square hole
through its centre. Scale *.
6- A rectangular slab of stone, 18" long, 12" broad, 3" thick, with
a circular hole 5" diameter through its centre. Scale
7. The frustum of a cone, the diameters of the ends 2.5" and 1. 5",
length 1 ".
8. A square pyramid, base 2^" side, axis 3V long.
9. A hexagonal pyramid, base 1.5" side, axis 2.5" long.
10. The frustum of the pyramid, Ex. 9, obtained by a plane bisect-
ing its axis at right angles.
11. A hemisphere, 3. 1" diameter.
12. A hemispherical bowl, 20" diameter inside, J" thick. Scale |.
13- A semicircle, and a quarter circle, 3" diameter.
14- A half cone, 2.5" base, 2" axis, obtained by a cutting plane
through the axis of the cone.
15. A half cylinder, 2" base, 1.5" long, obtained by a plane con-
taining the axis.
16. A quarter cone and a quarter cylinder, of dimensions as in
Exs. 14 and 15, obtained by planes containing the axis.
17- A quarter sphere, and an eighth of a sphere, obtained by two
or three mutually perpendicular planes through the centre of a
sphere 4" diameter.
18. A cone, base i\" diameter, axis 1^-", resting with its base con-
centrically on one end of a cylinder, i^" diameter, 2" long.
19- A ring, 2" internal diameter, of section -J" square.
20- A ring, 2" internal diameter, of circular section, \" diameter.
21- A regular tetrahedron of 3" edge.
22. A regular octahedron of 2" edge.
Hint. In examples like 21 and 22 there is a choice as to
which three perpendicular lines connected with the solid shall be
taken for the three principal directions. In the tetrahedron we
might select the base and altitude of one triangular face, and the
direction perpendicular to the face. In the octahedron the three
mutually perpendicular diagonals of the solid might be chosen.
23. A cone of indefinite length having a vertical angle of 50 .
24. The surface of revolution, page 337, double the size shown.
Hint. Employ inscribed spheres in Exs. 23 and 24.
Note. In the above examples the student may use either an
isometric or an ordinary scale in setting off dimensions in the
isometric views.
452 PRACTICAL SOLID GEOMETRY chap.
353. Problem. To determine the isometric projection
of the solid given in plan and elevation.
Draw the isometric axes a x x, a x y, a x z, and set off a x b v
a x d v and a x e x respectively equal to the isometric lengths (as
obtained by a scale) of the lines ab, ad, and a'e . In like
manner obtain the projections of the other edges parallel to
AE. The four arcs joining the upper extremities of these
projections must now be determined.
Divide the circular arc e'f into a number of equal parts,
say four. Join e x f x . Consider one of the points, say r ;
draw r'ri perpendicular to e'f . Set off e x n and nr x re-
spectively equal to the isometric lengths of e'ri and n'r ;
this determines r v and the corresponding point on the arc
parallel to e x r x f x is obtained by drawing r x r x equal and
parallel to a~d v By a repetition of this construction the
four arcs may be obtained.
To determine the isometric projection of the cylindrical
portion of the solid we may proceed as directed in the
last problem or as follows : Suppose that the cylinder is
circumscribed by a square prism which extends to the
base of the solid ; its plan and elevation are shown in
the figure, and its isometric projection may be thus ob-
tained
Draw the diagonal b x d v and from its middle point i\ set
off i x t x and i x u x respectively equal to it and in. ( TU is
not shortened by projection.)
Make i x w x and i x v x respectively equal to the isometric
lengths of i'w and i'v. It will then be seen that the three
parallelograms with q, w v and v x for their centres can be
completed.
The ellipses which are the projections of the circular
ends of the cylindrical portion of the solid must now be
inscribed in the parallelograms of which w x and v x are the
centres. Make w x g equal to the true length of the radius
of the cylinder, and complete both ellipses in the manner
explained in Prob. 352 ; two common tangents to the
ellipses will complete the required projection of the solid.
XVIII
METRIC PROTECTION
45 J
C
a'
d
I
IV
a
r
b
Examples. Represent in isometric projection
1. A sphere of 3" diameter penetrated by a cone of base 3" diameter,
axis 4" long, the centre of the sphere being at the middle point
of the axis of the cone.
2. The wedge-shaped figure formed from a cylinder 3" diameter,
3" long, by two sloping planes containing a diameter of one
end, and touching the circle at the other end at opposite points,
thus cutting away the sides.
The instrument box of Ex. 6, p. 447, with the lid opened at
3.
130
Scale
\ or '
4. One of the four quarters of a hollow sphere, formed by two
perpendicular planes containing a diameter. External and
internal diameters 3" and 2".
5. The separated parts (V) and (</) of the dovetailed joint, p. 457,
represented (fitted together) by ordinary projection at (a) and
6. The bolt with hexagonal head, p. 425, to double the size shown.
The upper surface of the head is spherical.
7. The bolts with round and square heads, p. 425, to the dimen-
sions given. The upper surface of the square head is spherical.
Note. In working these examples, the use of an isometric scale is
optional.
454 PRACTICAL SOLID GEOMETRY CHAP.
354. Problem. Having given a set of trimetric
axes, to determine the corresponding trimetric scales.
Let ox, oy, oz to the left be the given axes. Take any
point d in oy, and draw df, fe, ed respectively perpendicu-
lar to ox, oy, oz. Now ox, oy, oz are the projections of
three lines OX, O Y, OZ in space mutually perpendicular,
and df, fe, ed are the traces of the planes YZ, ZX, XY on
the plane of projection, or on a parallel plane.
On ed as diameter, describe a semicircle cutting oz in o Q ;
join o e, o Q d. Thus eo Q d is a right angle, and the triangle
eo d is the rabatment of EOD into the horizontal position
about ed. Thus eo Q and do are the true lengths of the
lines of which eo, do are the metric projections.
Again, the right-angled triangle of which the line fgo is
the plan, is shown rabatted at fgo l ; so that fo x is the true
length of the line of which fo is the metric length.
The construction for the trimetric scales is thus at once
obtained. Draw Ox, Oy, Oz parallel to the given axes, and
draw OX, OY, OZ respectively parallel to o Q e, o () d, o x f. On
OX, OY, OZ set off true length scales, and project these
scales on Ox, Oy, Oz by lines parallel to oo rj oo , oo v
The scales thus projected on Ox, Oy, Oz are the
trimetric scales required.
Note. Having given the trimetric scales 1, in, n, or their ratios
I : in : n, to determine the trimetric axes.
The values /, m, n are the ratios of the projected to the true lengths.
Construct a triangle xyz, the sides yz, zx, xy of which are respect-
ively equal or proportional to / 2 , nfi, and n 2 . Determine o the
centre of the circle inscribed in this triangle, and join ox, oy, and oz ;
these three lines are the required trimetric axes.
It can be shown that P + in- + n- = 2.
Examples. 1. The angles xoz xa&yoz between given trimetric axes
are respectively 135 and 120. Construct the trimetric scales.
Measure the representative fractions of these scales.
.4ns. I, in, n for the scales ox, oy, oz are .S38, .932, .649.
Note. The trimetric axes in this example may be drawn with the
90 , 45 , and 30 edges of the set-squares.
2. If the trimetric scales /, in, n are in the ratios I : J : |, determine
the trimetric axes and measure the angles between them.
Ans. lof, 138.5, 114.5-
XVIII
METRIC PROJECTION
45!
Examples on Problems 354 to 357.
Represent in trimetric projection the following six objects, the metric
axes being such that oz is vertical on the drawing-paper, the
angle xoz 135 , and the angle yoz 120'. All the lines maybe
drawn with the ordinary set-squares.
1. A cube of 3" edge, with a circle inscribed in each visible face.
2. A building brick 9" x 4^" x 3". Scale \,
3. A sphere of 2" diameter.
4. Three lines, each 3" long, which mutually bisect one another at
right angles.
5. The rivet on p. 427 to double the size shown.
6. The instrument box of Ex. 6, p. 447, with the lid open at 120 .
7. Refer to Art. 357 and the figure on p. 459. Draw the axes oa,
ob, oc respectively horizontal, vertical, and with the 30 edge of
the set-square. Take the scales for lines parallel to oa and ob
full size, and the scale for lines parallel to oc half size. Then
draw the projection of a cube of 2.5" edge, with a circle in-
scribed in each visible face.
8. Take the axes and the scales as in Ex. 7, or alter them in any
manner that may seem desirable, and represent in metric pro-
jection the model or the three planes of reference as shown in
Eig. 161 (/>), making OX, O V, and OZ each 4".
In this view plot the point A, the line CD, and the plane
ABC of Ex. 11, p. 447.
9. Take the axes and scales as in Ex. 8, and draw the metric pro-
jection of the steps, Eig. (a), p. 461.
456 PRACTICAL SOLID GEOMETRY chap.
355. Problem. Having given the trimetric axes ox,
oy, oz, to draw the trimetric projection of a cube of given
edge with a circle inscribed in one face.
First determine the trimetric scales as in Prob. 354. The
axes and scales in Prob. 354 are used again.
To project the cube, set off the lengths oa, ob, oc for the
edges, each to its proper scale. Complete the figure by
drawing the parallel lines as in isometric projection.
Let the circle be inscribed in the right front vertical
face ; its projection is the principal ellipse inscribed in the
projection of this face. The ellipse may be set out as in
Prob. 88, or as follows :
Draw the two diagonals ac, of, and through their intersec-
tion / draw the two trimetric diameters gs, pr parallel to oy,
oz ; then /, g, r, s are four points in the ellipse. Draw the
separate figure, centre I, and then locate n by plotting the
co-ordinates om, mn equal to OM, MN reduced by the
scales. The other points ti, v, w on the diagonals are then
determined by drawing the projection of the inscribed
square. The ellipse can now be traced through the eight
points found. The tangents at all these points are known.
Note. Observe that the diameters itv, nw which lie on the dia-
gonals ac and of are not the major and minor axes of the ellipse, as was
the case with isometric projection. The major axis is the line through
7 perpendicular to ox, of length equal to the true diameter of the circle,
for this line is parallel to the plane of projection. And generally, a
line in or parallel to one face, say xoy, is parallel to the metric plane,
when its projection is at right angles to the perpendicular axis oz.
356. Problem. Two views (a) and (b) of a dovetailed
joint are given. To draw a trimetric projection of each
portion of the joint, the trimetric axes and scales being
taken the same as in Prob. 354.
The required projections are shown at (c) and (d).
After the detailed descriptions previously given, the method
of obtaining these should not require further explanation.
The lines EH and FL, not being parallel to a trimetric
axis, their projections cannot be scaled. Their ends are
located by the method of co-ordinates. See the line KH.
XVIII
METRIC PROJECTION
457
M
x kjf
X"
\
V
1 *'
\ /
><
c
s'
t'
0'
H
9'
"
f e r o
(a)
I h
(b)
356
458 PRACTICAL SOLID GEOMETRY chap.
357. Generalisation of the foregoing methods. The
object of metric projection is to define a form having three
dimensions by one pictorial view drawn to scale. In the
examples hitherto considered this has been effected by
orthogonal projection. We may, however, remove the restric-
tion that the projection shall be orthographic, and still
secure the object in view, and at the same time gain con-
siderably in freedom.
Radial projection is not permissible, because a perspective
view cannot be scaled ; but we may employ oblique parallel
projection, since in this case parallel lines project into parallel
lines, all to the same scale. Now if the projectors, instead
of being perpendicular, may be inclined, and in any
direction, we gain two degrees of freedom in arranging for
the projection. We may now select what two angles we
like for the latitude and longitude, or the altitude and
azimuth, so to speak, of our projectors.
Let us now see how this helps us. In trimetric ortho-
gonal projection we may choose the axes, and then we
require to determine the corresponding scales. In trimetric
oblique projection, with two more degrees of freedom, we
may choose the axes and any two of the scales ; or what is
equivalent, the axes and the ratios /:///: ;/ of the scales,
leaving the absolute scale to be determined by a geometrical
construction, if this should ever be required. But in
practical applications we are not concerned with the
absolute ratio which the size of the object bears to the size
of its projection ; thus in isometric projection we generally
use the full size instead of the isometric scale, and do not
trouble ourselves as to how much bigger the object would
have to be in order to actually project into the view drawn.
Discarding consideration for the absolute scale, the general
proposition may be thus stated :
Proposition. /// trimetric oblique projection we may take
the three axes in any directions we like, and 7ve max take
the three scales whatever we like, and we shall have a true
projection of the object, or of one similar in form.
XVIII
METRIC PROJECTION
459
As a simple illustration, draw oa, ob equal and perpen-
dicular to one another, and draw oc in any direction and of
any length. Complete the figure
as shown by drawing the series of
parallel lines. Then we are at
liberty with perfect propriety to
take this figure as the projection
of a cube. Thus let the cube
rest on the plane of projection,
with one face coinciding with
oadb. The line oc is the projec-
tion of the perpendicular edge oC.
Therefore Cc must be one projector, and the oblique pro-
jectors are parallel to Cc. But generally it would not be
so easy as in this case to locate the position of the object
in space in regard to its projection. We may, however,
employ the projection without solving this latter problem.
We may remove another restriction. The three-
directional system of lines in space need not be perpen-
dicular to one another, but may have any directions. The
proposition in its most general form may then be stated :
Theorem. Let OA, OB, OC be any three lines of definite
length in space, and oa, oh, oc any three lines of definite length,
in one plane : then the former lines ran be projected into a
figure similar to the latter by parallel projection.
In applying this general method to practical cases, some
regard must be had to the effects of distortion. Any pro-
jection would appear right if viewed in a direction parallel
to the oblique projectors. But a picture is generally looked
at from somewhere near the front ; if the projectors were
very oblique this front view would appear very distorted; so
the axes and scales must be kept within reasonable limits.
As in ordinary trimetric projection we can plot the pro-
jections of irregularly situated points or lines from their
co-ordinates, and circles may be plotted by projecting the
circumscribing square or parallelogram, and drawing the
principal inscribed adlipse.
46o PRACTICAL SOLID GEOMETRY chap.
358. Miscellaneous Examples.
*1. Fig. (a). Draw the isometric projection of the object represented
orthographically. A section on AB is shown. Scale T ^.
(1896)
*2. Fig. (b). Draw the letter P from the dimensions given in the
sketch. Assuming the thickness of the material from which it
is cut to be y", make an isometric view of the letter.
N.B. An isometric scale is not to be used. (1S93)
*3. Fig. (t). Make an isometric view of the cross. {^19)
*4. Fig. (d). Make an isometric view of the bolt head, the elevation
and half plan of which are given.
N.B. An isometric scale is not to be used. Lengths to be
transferred direct from the given figure to the isometric lines.
(1886)
*5. Fig. (e). The two lines ad, ac form one of the right angles of a
face of a cube of 2|" edge. Complete the plan of this cube.
Unit o. 1". (1 89 1)
*6. Fig. (/). Two elevations of a gable cross are given. Make an
isometric view of the cross. An isometric scale is not to be
used. (1892)
7. Make an isometric view of the bolt, Fig. {d), p. 441, drawing it
full size. An isometric scale is not to be used. (1887)
8. Particulars of a trestle are given in Ex. 13, p. 440. Represent
the trestle in isometric projection. Scale \. Use of isometric
scale is optional.
9. Draw an isometric projection of the desk shown in figure, p.
203. Use of scale optional.
10. On each face of a cube of 2" edge stands an equal cube. Make
an isometric view of the solid formed by these seven cubes. An
isometric scale is not to be used. (1878)
11. Draw the isometric projection of the object in Fig. (a), p. 441,
standing on the horizontal plane, one isometric plane to be
taken parallel to a diagonal such as fg, passing through two
opposite edges of the fluted column. An isometric scale is not
to be used. (1897)
12. Draw three lines meeting at a point and making angles of 120,
1 30, no Q , with each other. These lines are the plans of the
edges of a cube of 3" edge. Complete the plan of the cube and
draw its elevation on any plane parallel to no side or diagonal.
(1879)
X V 1 1 1
METRIC PROJECTION
461
Copy die figures dvub/e sc^e
y
\
(a)
/ is'
-6'-A
i
1 1
^r
+ ]
ix
CHAPTER XIX
MISCELLANEOUS PROBLEMS
359. The five regular polyhedra. It is shown in
treatises on pure solid geometry that there are five, and
only five regular polyhedra. These are the tetrahedron,
cube, octahedron, dodecahedron, and icosahedron ; see the
Appendix, Definitions 18 to 22. We have already had
occasion to represent the first three in projection ; the
remaining two will now be considered.
The regular dodecahedron, Fig. (1). This solid has
twelve equal and regular pentagonal faces. The projection
easiest to determine is that on the plane of one face.
To obtain this, draw the regular pentagon aaaaa to re-
present the bottom face. On two adjacent sides construct
regular pentagons, and regard these as the rabatments of
two adjacent faces. Thus aB , aB are the rabatments of
the same edge about the two axes aa, aa. From B w B
draw perpendiculars to the axes intersecting in b ; then
ab is the plan of a sloping edge. The plan may now be
completed from considerations of symmetry
From the corners a, a, a, a draw the plans ab of the
other four similarly-situated edges AB. For the top face,
draw the regular pentagon ddddd, with sides parallel to aaaaa
and circumscribing: the same circle. Draw the five lines
dc each equal to ab, to represent the five edges sloping
from the top face. Complete the plan of the solid by
drawing the outline, which is a regular decagon.
chap, xix MISCELLANEOUS PROBLEMS
S B,
4"3
The distances of the points B, C, and D from the lower face A are
readily found since we have the plans, and we know the true lengths of
the lines AR, BC, and CD : we can thus draw an elevation on any
vertical plane.
If we draw an elevation on a plane parallel to a diagonal AD of the
solid (a diagonal being a line which joins opposite corners), then we can
measure the length of the diagonal ; and from the elevation we may
project a plan with the diagonal vertical. This plan might also be
drawn first-hand from considerations of symmetry.
The regular icosahedro/i, Fig. (2). This solid is bounded
by twenty equilateral triangular faces. Each angular point
of the solid is the common vertex of five triangles, the
bases of the triangles forming regular pentagons. We shall
obtain the plan of the solid when a diagonal is vertical.
1 )raw two regular pentagons b, b . . . c, c . . . circum-
scribing the same circle and with sides parallel. Join all
the points b and c to the centre, as shown in the figure, and
also join the adjacent points b and c so as to obtain the
regular decagonal outline. This figure is the projection of
the solid on a plane perpendicular to the diagonal AD.
We may obtain the heights of the various points, the length of the
diagonal, the distance between parallel faces, and the projection on the
plane of a face, by methods suggested for the dodecahedron.
464 PRACTICAL SOLID GEOMETRY chap.
360. Problem. To determine the spheres inscribed in
and circumscribing a given regular polyhedron.
Let the given solid be a regular tetrahedron, of which
abed in the plan.
The inscribed sphere. General method. First draw a
view of the solid on a plane perpendicular to an edge, so as
to project two adjacent faces in profile. Next obtain the
projection d of the centre of the solid. Finally with centre
d draw the circle which touches the profile projections of
the two faces. This will be the projection of the inscribed
sphere.
Thus take xy perpendicular to ab, and draw the elevation
a'b'c'd'. Draw the perpendicular d ' m and the bisector an
to intersect in d ' . With centres d and o, radius dm', draw
two circles.
These circles are the projections of the inscribed sphere.
The circumscribing sphere. General method. First
obtain the projection of the solid on a plane equidistant
from the centre O and two corners. Then with centre d
draw the circle through the projections of the corners.
This will be projection of the circumscribing sphere.
Take ay parallel to ocd, and draw the elevation a'b'c'd.
Obtain d as before. With centres d and o, radius o'd\ draw
two circles.
These circles are the projections of the circumscribing
sphere.
Note. To determine the spheres inscribed in, and circum-
scribing a given irregular tetrahedron.
For the inscribed sphere, first determine the three planes which bisect
any three dihedral angles of the solid. Then determine the sphere
which has its centre at the point of intersection of the planes, and
which touches any face of the solid.
For the circumscribing sphere first obtain the three planes which
bkect at right angles any three edges of the solid. Then determine the
sphere which has its centre at the point of intersection of the planes,
and passes through any coiner of the solid.
It is seen that this problem is the same as that of finding a sphere
which shall touch four given planes, or contain four given points.
XIX
MISCELLANEOUS PROBLEMS
465
361. Problem. To determine any regular polyhedron
inscribed in or circumscribing a given sphere.
Let the circle, centre o v be a projection of the given
sphere, and let the required figure be a tetrahedron.
First work the last problem for a tetrahedron of any
assumed edge, that is, draw Fig. 360.
Through o x draw lines parallel to d'm' and a'n .
The inscribed tetrahedron. Through d x draw d x a v d x c x
parallel to d'd, d'/, and join c x a v which will be parallel to
c a .
The circumscribing tetrahedron.- Draw the tangents a.^d.,
a.,c, parallel to ad', dc, and through d., draw d c.> parallel 10
d'c'.
The lengths of the edges of the inscribed and circum-
scribing tetrahedra are d x c x and d./., respectively.
A similar method is used for any regular solid.
2 H
466 PRACTICAL SOLID GEOMETRY chap.
362. Trihedral angles and spherical triangles. Let
O be the apex of any trihedral angle, and suppose a section
of the angle to be made by any
-B/ spherical surface whose centre
s^\\ i s at O- 1 ne sections of the
s' \ three faces will be arcs of ereat
s' \ /y circles, forming a spherical tri-
/^l_ J 4 angle ABC on the surface of
^^^^ / / the sphere.
^"~~\/ We may let A", B, C stand
^4""^ for the angles, and a, b } c for
the sides. By the former we
mean the angles between the tangents to the curved sides
at the corners, and by the latter the angles subtended by the
curved sides at the centre of the sphere. Thus the angles
A, B, C\ and the sides a , b, c" of the spherical triangle
are respectively equal to the three dihedral angles and
the three plane angles or faces of the solid trihedral angle.
The polar triangle. If through the centre of the sphere, lines OA v
OB v OC x be let fall respectively perpendicular to the faces OBC, OCA,
OAB, and all directed outwards (or all inwards), meeting the spherical
surface in A v B v C v we have a second trihedral angle formed, with its
corresponding spherical triangle A,B^C,. The latter is called the
polar triangle of the triangle ABC.
We have not space to establish the well-known relations between
the sides and angles of the triangles ABC, A l B l C v therefore we shall
merely state them. Denoting the sides and angles of the polar triangle
A l B l C l by tfj , /; t , c x ; A, B^, Cj, it may be shown that the angles
of the one triangle are supplementary to the sides of the other ; that is,
we have
<4 1 B =l&o a -a a l =i8o -A
B^ = i8o" - i> b 1 =i8o-B
C l =iSo-e * 1 =i8o -C
The relations between the two triangles are reciprocal; each triangle
is the polar of the other.
The principal use of the polar triangle is in the solution of spherical
triangles. It reduces, to one-half, the number of cases necessary to be
dealt with.
Problems on spherical triangles are equivalent to problems on tri-
hedral angles. In solving problems on the former we shall require to
draw projections of the latter.
XIX
MISCELLANEOUS PROBLEMS
467
363. Problem. To solve a spherical triangle or tri-
hedral angle, having given : I. Three sides, or three angles.
II. Two sides and the included angle, or two angles and the
adjacent side. III. Two sides and one angle opposite to
one of them, or two angles and a side opposite to one.
Case I. Let the three given sides be a", //', c.
Begin by drawing a development of the trihedral angle.
That is, draw any circle, centre O, and set off the angles
B x OC, CO A, AOB x equal respectively to the given sides
a', />', c. Through B v B x draw perpendiculars to OC, OA
intersecting in /; ; then b is the plan of the point B when
the two outer sides or faces a and c of the solid angle are
turned into their true positions, about <9Cand OA.
Join Ob. We have now obtained a projection of the
solid angle on the face OAC. The projection AbC of
the spherical triangle is also shown. Several simple methods
of setting out the elliptic arcs Ab, ^Cwill suggest themselves
to the student.
Through B v B x draw tangents to the circle, that is
468 PRACTICAL SOLID GEOMETRY chaf.
perpendiculars to OB v 0B V meeting OC and OA pro-
duced in //and K.
The three required angles can now be found. The
angle B is obtained by the rabatment HB ? Kol the triangle
HBK. The angles A and C are equal" to bNB^ bMB 2 ,
determined by the rabatments of the triangles bNB, bMB.
The problem is thus solved.
If the three angles A, B, C were given, the problem
could be reduced to the one just worked, by means of the
polar triangle. We should first subtract each angle from
i8o, and thus obtain the sides of the polar triangle.
Then by the construction above we could find the angles
of the polar triangle, the supplements of which would give
us the required sides of the original triangle.
Note I. Observe the properties of the figure.
There are five rabatments of the point B about five different axes,
from which we have the relations
MB l ^MB 2 ; NB x = NB t \ &B 2 = bB 2 ; HB X =HB Z ; KB X =KB % .
Since the plane HBK'xs, perpendicular to the line OB, the trace HK
is perpendicular to the projection Ob. Hence also B :i falls on Ob,
Note 2. A simple and effective model can be made by cutting out
the shape OB^HKB^O in paper, drawing on it the arc B\CAB V and
indenting and folding along OH, OK. The three triangles Hk'B v
bJI/B.,, bNB., are also cut in paper, with margins along their bases for
attachment to the model by glue or paper-fasteners.
Case II Let the given sides be a, />', and the included
angle C.
First draw the rabatted sides B x OC, COA equal to
a , b. Through B x draw a perpendicular to OC, intersect-
ing the latter in M. Set off the angle bMB 2 equal to C\
and make MB 2 = MB V Through B draw Bb perpen-
dicular to BM produced. Then b is the plan of B.
Through b draw 1>NB 1 perpendicular to OA. We thus
find the side c, and the angles A" and B are readily
determined by rabatments.
If we are given two angles and an adjacent side this can
be at once reduced to the above by means of the polar
triangle. Or the problem can be easily worked directly.
XIX
MISCELLANEOUS PROBLEMS
469
Case III. Let a, b , A be the given sides and angle.
First draw B\OC, CO A the development of the two
given sides or faces.
Next set out the elliptic arc Ab, of indefinite extent,
which shall be the projection of the arc AB X when the
latter is turned about OA until the face OAB is inclined
at the given angle A to the plane OAC.
Then through B x on the side a draw a perpendicular
B X M to OC to intersect the elliptic arc in two points b, />.
The plan of B is thus found, and the solution can now
be completed as in the preceding cases. There are two
solutions, so that Case III. is ambiguous.
Note 3. The drawing of the elliptic arc may be avoided and ihe
points of intersection found by the method of Prob. 95.
The case where two angles and a side opposite to one
of them are given reduces at once to the present case by
means of the polar triangle.
47o PRACTICAL SOLID GEOMETRY chap.
364. Problem. It is required to fit a cylindrical shell
eccentrically on a hemispherical dome, as shown in the
figure. Draw the elevation of the intersection of the
surfaces, and develop the plate for the cylinder.
As the development is required, take equidistant gener-
ators on the cylinder, and find the points where these meet
the sphere.
Through c, a draw the diameter of the plan of the cylinder ;
divide the semicircles each into the same number (six) of
equal parts, figuring the points as shown.
Turn c$, c$ into the position c$ x parallel to xy; project
from 3j to 3 X '; draw the horizontal line through 3/ to meet
the projectors from 3, 3 in 3', 3'.
Repeat the construction for the other points, and draw a
fair curve through the points 1', 2', . . . thus found. This
is the required elevation of the intersection.
The development. On any line set off HAT, MK each
equal to \ the circumference of the cylinder, Prob. 113.
Divide HK into six equal parts. Erect perpendiculars of
lengths M^ 1 = m'^', J\ r 2 1 = 112', . . . Draw a fair curve
through the points o,, 1,, 2 l5 . . . as shown.
The development of a symmetrical half of the cylindrical
shell plate is thus obtained.
365. Problem. To develop the surface of a sphere ap-
proximately (a) in zones ; (b) in lunes.
The centre of the sphere is taken in XY, and the vertical diameter
is the polar axis.
A zone is the surface included between two planes, perpendicular to
the axis ; as between two circles of latitude. A lime is the surface in-
cluded between two planes meeting along the axis ; as between two
semicircles of longitude.
(a) Divide the arc ric into three (or other number) of
equal parts. Draw da, b'b' . Join b'd, b'd and produce to
meet in v. Then the frustum AB of the cone VB coin-
cides approximately with the zone AB.
Develop this frustum. One half only is shown. Arc
b'B 1 = semicircle bsb. Prob. 113.
XIX
MISCELLANEOUS PROBLEMS
471
rrV ii' a'
~or^
>^
H
N\
\M \"
I4 6 t
X
The other zones must be similarly developed,
(b) Make // = a'b'. Join ot, ot. Then tot is the plan of
a lune of T V the surface.
Set off oN x arc ric
(Prob. 113).
Divide N x into
three equal parts by R X R V S X S V making R X R X rr, S X S X =
ss. Draw the curves t'S v R X N X as shown.
This is the development of the upper half of the lune.
Examples. 1. Develop approximately the surface of a hemi-
sphere (a) in four zones ; (b) in sixteen Junes.
2. Two copper pipes 10" diameter, with their axes at right
angles in one plane, are connected by an elbow pipe in the
form of a quarter annulus, the mean radius of which is 10''.
Develop approximately the plates for the elbow (a) in twelve
zones ; (b) in four equal lunes. Scale x
472 PRACTICAL SOLID GEOMETRY chap.
368. Problem. The shape of the surface of a piece
of ground is given by contour lines indexed in feet, and a
horizontal scale of feet. The indexed plan of a flat sur-
face ABCD is also given. The plot is formed partly by
cutting and partly by embankment, the slope of the former
being 45 and of the latter 40. It is required to draw
the contours and boundary of the completed earthwork.
The contours or lines of level on any inclined plane
surface are straight lines perpendicular to the line of slope.
Their distance apart in plan for any difference of level may
be found by the following simple construction :
Draw two intersecting lines, one vertical, the other
inclined at the angle of slope, say 40^'. From their inter-
section P set off PQ vertically to represent any difference
in level, and draw QR horizontally to meet the inclined
line in K. Then QR is the distance between the contour
lines in plan.
When, as in the above cutting, the slope is 45 , we have
QR = PQ, or the horizontal distance is equal to the vertical
distance ; the construction is not then required.
We shall begin the solution by determining the contours
at 5-feet intervals on the cutting which springs from AB.
With a. A2 , b zx as centres, draw circles respectively of 3-feet
and 4-feet radius to scale. The common tangent to these
circles is the 3 5-feet contour on the cutting. The other
contours are then drawn parallel to this line at distances of
5 feet. A curve drawn through the points marked 1, where
these contours intersect the given contours at the same
levels, gives the upper boundary of the cutting on this side.
Next consider the cutting which rises from AD.
With ^/., 9 as centre, radius 6 feet to scale, describe a
circle ; the tangent common to this circle and to the one
with a. i0 as centre is the 35-feet contour on this cutting.
The remaining contours are drawn parallel to the one
just found at intervals of 5 feet. Their points of intersec-
tion with the given contours are marked 2 in the figure.
The freehand curve through the points 2 is drawn, and
XXI
MISCELLANEOUS PROBLEMS
473
10 10 ZO 50
uilini
35
30
X
25 20
intersects the first curve in /and the edge of the plot in e.
These points represent the highest and lowest points of the
cutting on this side.
AF is the line of intersection of the two cuttings. At
the point E the embankment begins.
The student should now be able to complete the problem
without further detailed description.
For the embankments, which slope at 40 , the radii of the
circles and the distance apart of the contours are determined
by the horizontal lines of the scale PQR, as explained at
the beginning.
The line gh is the contour 30 across the plot ; BG
and DH are each one-third the length of the plot.
This is a good example of the use of figured plans.
474 PRACTICAL SOLID GEOMETRY chap.
367. Problem. A given parallelogram pqrs is the
projection of a certain square, determine the length of the
side and the inclination of the plane of the square.
Draw the bisectors^ and r
By Prob. 87 determine aa v bb v the major and minor
axes of the ellipse of which jj v ii x are conjugate diameters.
This ellipse will be the projection of the circle inscribed
in the required square, and the major axis is the projec-
tion of that diameter of the circle which is horizontal.
Thus aa x is the length of the side of the square.
Draw xy perpendicular to aa v and from b draw a
projector to meet, in b', an arc with c as centre and ca as
radius. Then the angle b'c'x is the required inclination of
the plane of the square.
The edge elevation of the square will be parallel to c'b',
and we can draw it when we know the height of some
point connected with the square.
The rabatment of the square about AA X into a horizon-
tal position is easily found if required.
368. Problem. A given triangle ahc is the projection
of a triangle ABC, the latter being similar to a given
triangle ABC; it is required to find the actual size of
ABC, and the inclination of its plane.
On any side, say B'C, of the given triangle A'B'C
describe a square ; join A' to one corner P' of this square,
and meeting B'C in R' .
In the corresponding side be obtain r such that
br:rc = B'R':R'C.
Join ar and produce to p such that
ar : rp = A'R' : R'F'.
Join bp and draw cq, pq parallel to bp, be.
Then beqp is the plan of the square attached to BC\ the
triangle ABC and the square BCQP lying in the same plane.
Determine as in the preceding problem the inclination
of this plane, then by a rabatment obtain the actual size of
the triangle ABC
XIX
MISCELLANEOUS PROBLEMS
475
c I
$ 368
Note I. The points r and p may readily be obtained by describing
a square on be instead of B'C, and constructing a triangle on be of the
same shape as A' B'C.
Note 2. The student should be able to extend the construction so
as to solve the following more general problem :
The plaits a, b, e are given of three known points A, B, C connected
with any plane or solid figure ; complete the plan of the figure.
476 PRACTICAL SOLID GEOMETRY chap.
369. Problem. To project a helix of given diameter
and pitch.
Definitions. A helix is traced on a cylinder when a point travels
round the surface at the same time advancing axially, the ratio of the
two speeds being constant. The advance per revolution is the pitch.
If a right-angled triangle ABC having the base AC= circumference,
and the height Ci? = pitch, were made in paper, and wrapped round the
cylinder, bringing the points A and C together, the hypothenuse would
become a helix. The base angle A is the pitch angle of the helix.
To obtain the projection, we proceed as in Prob. 130 for the sine
curve. The latter may be regarded as the projection of a helix.
Draw the semicircle, centre /, of the given diameter.
Project cc of length equal to the given pitch.
Divide the semicircle into six (or other number) of equal
parts from o to 6' ; and divide cc into double the number
by the perpendicular lines o to 12.
Project from o', i', . . . on the lines 0,1,... Then
draw the helical curve through the points as shown.
Note. A second turn of the helix, with the projection of a helical
spring of circular section, is shown. This is determined as the envelope of
the projection of a sphere which maybe assumed to generate the surface.
370. Problem. To project a screw thread and a spiral
spring of square section.
Definition. If the generating point above be replaced by a line,
the helix becomes a helical surface. The line or figure tracing the
helical surface may be conceived as attached to a screw which turns in
a fixed nut.
A line perpendicular to the axis, shown in the various
positions, 00, 11, 22 . . . generates one of the helical
sides of the screw thread. The spiral spring may be re-
garded as having been traced by the square s.
The pitch cc is the same as before, and the divisions o
to 1 2 are again made use of in projecting the screw.
371. Problem. To project a right-handed V threaded
screw and a longitudinal section of the nut.
After what has been said above, the construction should
be clear from the figure, without description.
The screw is single threaded, and the pitch pp.
Note that a line such as act is perpendicular to the axis, since in a
half-turn the thread advances half the pitch.
XIX
MISCELLANEOUS PROBLEMS
477
0'
I\
r
r^o 1
12-
i 2 '
-J- 1
\j
f
4'
478 PRACTICAL SOLID GEOMETRY chap.
372. Problem. A ruled surface is generated by the
tangent moving along the given vertical helix ABCD.
(a) Draw the horizontal trace of the surface, (b) Obtain
the elevation of the section made by the given vertical
plane LM. (c) Show the envelope of the tangent in plan
and elevation.
The point A is on the ground, and aoc is perpendicular
to xy.
Draw the tangent at c, and by Prob. 113 set off ck = arc
cb. Make ch = twice ck = rba ha\( circumference. Pro-
ject from h to //. Join and produce h'c .
Then HC is the tangent to the helix at C. See Prob. 369. The
base angle a of the right-angled triangle AIIC is the pitch angle of the
curve, and all the tangents are inclined at a to the ground. If the
triangle AHC were wrapped round the cylinder containing the helix,
the hypothenuse (produced) would generate the required surface, and
the locus of H would be the horizontal trace of the surface. The latter
is thus seen to be the involute of the plan abed of the cylinder ; this
could be set out as in Prob. 125, or as follows :
(a) Draw the tangent at any point d in plan. Project
from d to d\ and draw the horizontal d'f to meet h'c in
f. Draw the vertical _/; Mark off de equal to gk\ Pro-
ject from e to e and join d'e.
Then DE is the tangent to the helix at D, and E is its
horizontal trace. Find similarly other points in the required
trace of the surface, and draw the curve ahe through them.
(b) Obtain the points where the plane ZJ/cuts the tan-
gents like DE previously found. Or proceed thus :
Select any point p in Im. Draw the two tangents qp,
np. Draw the triangle with base angle a, and along its base
set off H y N v H X Q X equal to pu, pq. Erect the perpendicu-
lars dV l P v Q X P V and mark off their lengths as heights above
xy on the projector from p, thus obtaining/', p'.
Determine other points similar to p', p', and draw the
required elevation of the section through them as shown.
(c) The envelope in plan is seen to be the circle abed;
and in elevation is the curve a'b'c'd ' .
XIX
MISCELLANEOUS PROBLEMS
479
Thus the envelope is the helix itself. The surface has
an edge, which coincides with
the helix.
The figure shows how a model of
this surface may be made.
Set out the symmetrical curves
on cardboard, and make holes at
the points determined by a series of
equiangular tangents. Indent and
fold along the lines shown, and
secure by paper-fasteners through
A A, BB.
Use twine for the generators,
laced through the holes.
Take the dimensions of Ex. 27, p,
B
I C\ 6 J
to KJj?
A
pitch
B
f -?o
A
"V-p^
481.
480 PRACTICAL SOLID GEOMETRY CHAR
373. Examples on Problems 359 to 372.
1. Draw the plan of a regular dodecahedron of i" edge (a) when
a face is horizontal ; (b) when a diagonal of the solid is
vertical. Measure the length of the diagonal, the distance
between two parallel faces, and the distance between two
parallel edges.
2. Draw the plan of a regular icosahedron of i" edge (a) when a
diagonal of the solid is vertical ; (/>) when a face is horizontal.
Measure the length of the diagonal, the distance between two
parallel faces, and the distance between two parallel edges.
3. Determine the dihedral angles between the faces of a regular
dodecahedron and icosahedron. Find also the angles sub-
tended by the sides at the centres of the solids.
4. Determine the spheres inscribed in, and circumscribing, a regu-
lar tetrahedron, octahedron, and a cube, each of 2" edge.
5. Determine the spheres inscribed in, and circumscribing, the solids
of Exs. 1 and 2.
6. Find the sizes of the five regular polyhedra inscribed in, and
circumscribing, a sphere of 2" diameter.
7. The lengths of the six edges of an irregular tetrahedron ABCD
are respectively BC = 3", CA = 2", AB = 2\" , AD = 2%", BD =
2J?", CD=2\"- Draw the plan of the solid when the face
ABC rests on the ground, and index the plan of D.
8. Determine the indexed plans of the spheres which are inscribed
in, and which circumscribe, the irregular tetrahedron of Ex. 7.
Solve the following three spherical triangles :
9. Given a = 47, 3 = 55, c = 4l.
Ans. A" = 62, B = 81. 5 , C = 52-4.
10. Given ff = 75, 6 = 59, C = 5o.
Am. A = g7.s, B = 6i.6, ^ = 48.3.
11. Given a = 55 , b a = 8o, A" = 45 '.
Aits. c* = 6f, B =.$S.2, C =i27\ 4 ;
or _ ^==39.1, ^=i2i.8, C = 33-
12. Taking the answers in the above three examples as data, solve
the four triangles.
13. The latitude and longitude of Rome are respectively 41.9 N.
and 12.5 E., and of St. Petersburg 6o.o N. and 30.3 E. :
find the angle subtended by the two places at the centre of
the earth ; find also the direction at each place of the great
circle connecting the two. Ans. 25.05, 140.8; 21. 2.
14. Two copper pipes each 10" diameter meet at right angles to
form an elbow joint. Draw the development of the plate near
the joint for one of the pipes. Scale V'.
15. A horizontal range of sheet metal piping, 6"diameter, is required
to have a horizontal 4" branch fitted at an angle of 6o, the
xix MISCELLANEOUS PROBLEMS 481
upper surfaces of both pipes being at the same level. Draw
the development of the joint for both pipes. Scale ^.
16. Work Prob. 365, the diameter of the sphere being 4". Take
four zones in the hemisphere and develop them all. Let the
angle of a lune be 22^.
17. A square with one corner on the ground projects into a par-
allelogram, with adjacent sides 3" and 2", and included angle 6o\
Determine the side of the square, the trace and the inclination
of its plane, and the heights of its other corners.
18. Draw a parallelogram abed, taking ab = . 7", ad=^', 6ad=5$.
This figure is the projection of a rectangle whose sides AD,
AB are in the ratio of 3 to 2. Determine the sides and the
inclination of the plane of the rectangle.
19. An equilateral triangle of 2" side is the projection of a right-
angled triangle whose three sides are in the ratios 3:4:5;
determine the size and angular position of this latter triangle.
20. A right-angled triangle whose three sides are i\" , 2", 2j" is
the projection of an equilateral triangle ; determine the size
and angular position of the latter.
21. Draw a triangle oab making oa = 2", ob 1. 7", ab = 1. 3". The
point is the plan of the centre, and the line ab that of one
side of a regular octagon ; complete the plan of the figure.
Special solution. Draw any regular octagon ABC . . . centre O.
Join A C to intersect OB in M. In the given triangle take ;//
in ob such that om : tub = OM ': RIB ; join am and produce to
c, making mc am. The student should be able to complete
the solution.
22. In a regular tetrahedron of 3" edge take points P, Q, R on three
edges distant respectively 0.6", 1.5', and 2.2" from one corner.
Draw the plan of the solid when the triangle PQR projects into
an equilateral triangle.
23. Draw a triangle abe, have abT,", ae = 2", and 6e=l^"; ab is
the plan of one edge of a regular tetrahedron, and e that of the
middle point of the edge opposite to AB. Complete the plan
of the tetrahedron.
24. Work Prob. 369, having given: diameter=3"; pitch = 4";
diameter of section of springs 1".
25. Work Prob. 370, having given: diameters = 4" and 3" ; pitch
4" ; width of thread = -|".
26. Work Prob. 371, having given : diameter outside = 4"; angle
of thread = 6o ; pitch = J" ; length of nut = 4".
27. Work Prob. 372, having given : diameter of helix = 1"; pitch
= 2^" ; lm makes 15 with xy and touches the circular plan ac.
Make a model of this surface as described in the problem, using
stout cardboard.
2 I
482 PRACTICAL SOLID GEOMETRY chap.
374. Miscellaneous Examples.
1. A regular tetrahedron of 2" side is inscribed in a sphere. Draw
the elevation of the solids, the circumscribed sphere resting on
the horizontal plane, one face of the tetrahedron horizontal, and
one edge of that face inclined at 20 to the vertical plane.
f (i897)
2. Draw a triangle oab, having ab = .gz" and oa = ob = .$8". Take
as the plan of the centre of a sphere, radius ", and a, b as
the plans of two points on its upper surface. Determine the
plan of a regular triangular pyramid, circumscribing the sphere,
and having two of its sloping faces touching the sphere at A and
B respectively.
3. An octahedron of 2" edge stands on the horizontal plane. Draw
its plan and also that of its inscribed sphere. (1884)
4. Draw the plan of a cube inscribed in a sphere of 3" diameter
resting on the horizontal plane, one corner of the cube to be at
a height of 2.75" above the horizontal plane. (1878)
*5. Eig. (a). Determine the centre and radius of the sphere circum-
scribing the irregular pyramid ABCD. Unit = 0.1". (1882)
" ;; "6. Fig. (e). The three given parallel lines are the plans of portions
of the edges of a triangular prism. Determine the plan of a
sphere inscribed in the prism, and having 16 as the index of
its centre. Unit = 0.1". (1887)
*7. Fig. (d). The arc be is part of the plan of the base o. a right
cone standing on the horizontal plane ; v u is the plan of the
vertex. The portion ABCD is to be covered with paper.
Determine the shape to which the paper must be cut. Unit
= 0.1". (1888)
*8. Fig. (b). The end elevation and a portion of the plan of two
adjacent ridge roofs is given. Determine in plan the shortest
distance measured on the roof surface from A to B. (1887)
*9. Fig. (e). Determine the shape of the plates used to form the
elbow pipe with bell-mouth shown. Draw the figure four
times the size shown. (1887)
10. Fig. (f). Draw the development of the funnel shown, so as
to give the shapes of the plates from which it is made. Draw
the figure four times the size shown.
11. Draw a rectangle abed (ab = cd 3^" ; be = da=i^"); ah and cd
are the plans of the diagonals of two opposite faces of a cube ;
be and da are the plans of two edges. Complete the plan of the
cube. (1880)
12. Two equal right cones, height i|", diameter of base 2-3", have
a common vertex. One rolls upon the other which stands on
the ground. Determine the locus of a point on the circum-
ference of the base of the rolling cone. (Honours, 1SS6)
XIX
MISCELLANEOUS PROBLEMS
483
484 PRACTICAL SOLID GEOMETRY chap.
*12. Fig. (a). The area abed on a hill side, the traces of the plane
of which are given, is to be levelled ; the side slopes are I in 1 .
Complete the plan of the excavation. (1885)
*13. Fig. (/). The plan and section of an embankment are given ;
the lines aa, bb represent the sides of a road cut through it.
The slopes of the sides of the cutting are 35. Complete the
plan. ' (1884)
14. On a right circular cylinder of i|" diameter, a helix of 3"
pitch is traced. Draw the elevation of one turn of the helix :
draw also the plans and elevations of tangents to the helix at 1 2
equidistant points, and determine the points in which these
tangents meet the horizontal plane through the lowest point of
the helix. Axis of cylinder vertical. (1890)
15. A spiral spring, axis vertical, is of the form of a square screw
thread. Side of square V : external diameter on plan 3 " :
pitch 2|". Draw the elevation of one complete turn of the
spring. 11877)
16. A right cone, height 4", diameter of base 3", stands on the
horizontal plane. A point starting from the base of the cone
moves round its surface at a uniform speed, and rises half the
height of the cone in turning round to cut the generatrix from
which it started. Draw the plan and an elevation of the curve
traced by the point. (1SS6)
"*17. Fig. [e). A twisted surface of revolution is generated by a line
rft, a'c', revolving round a vertical axis 0, 00, to which it is
rigidly fixed by the horizontal line ob (in elevation />'). Draw
in plan and elevation the generating line when it has revolved
round the axis so as to pass in plan through/. (1894)
*18. Fig. (<). Two lines are given by their figured plans ab, cd. A
surface is generated by a line moving parallel to the horizontal
plane and always meeting both the given lines. Determine the
true form of the section of this surface by a vertical plane LM.
The section to be carried up to a height of 3" above the ground.
AYhat is this surface termed, and can it be developed? (1879)
19. Fig. (</). Two views of a crank are given. Determine the
proper shape of the dotted curve in the right-hand view.
20. A line 3" long moves at a uniform speed round a vertical axis, by
which it is bisected, and to which it is always at right angles. At
the same time it moves along the axis a distance of 1 |" for every
complete revolution, thus generating a helical surface. Make a
section of this surface by a plane parallel to and J" from the fixed
axis, the section to show two revolutions of the surface. (18S9)
21. Fig. (/"). A surface of revolution (a capstan) is shown in
elevation ; determine a sectional elevation on a vertical plane
i" from the axis.
XIX
MISCELLANEOUS PROBLEMS
4S5
Copy tfie figures double size- .
IV
3
i0h*&
- ^
~~ 7 u
C o
\o'
'e'
r -7
Xb*
(e)
a'/
\o'
U
CI
e
<0 p
SECTION III
GRAPHICS
CHAPTER XX
GRAPHIC ARITHMETIC
375. Graphic representation of magnitude. In ordinary
arithmetic we are concerned with numbers, and with numeri-
cal calculations. The former may be pure numbers, or they
may be concrete, that is, may express the magnitudes of
quantities of some specified kind. In the latter case the
number measures the magnitude by telling us how many
times the quantity contains an arbitrarily chosen unit of the
same kind and of known size.
In graphic arithmetic, numbers and magnitudes are re-
presented by the lengths of straight lines, set out or
measured to scale ; and calculations are made by drawing.
Thus let it be required to represent the number 7.5.
We may first draw any line, setting off any convenient
length on it to represent unity, and then on the same or
any other line mark off a segment 7.5 units long.
Or instead of drawing the unit line, we may set out or
specify some convenient scale, say J" to 1 unit, and using
this scale mark off the segment 7.5 units long.
If the number had been concrete, say 7.5 tons, then the
unit line would have been labelled 1 ton, or the scale would
have specified |" to 1 ton.
chap, xx GRAPHIC ARITHMETIC 4S7
II the number to be represented is comparatively large or small, as
for example 750 or .075, then we may replace the unit line by a line
which shall represent some convenient known multiple, such as 100
units, or submultiple like .01 unit. The scales are also correspond-
ingly modified ; thus we may take as suitable scales for plotting these
numbers ^" to 100 units, or V to .01 unit.
The student should refer to Probs. 7 to 1 1 on scales. Also to Art.
5 for the description of an engine-divided decimal scale, by the use of
which we are enabled to pass from the numerical to the graphical
system or vice versa.
376. Addition and subtraction. Lines to be added or
subtracted must represent quantities of like kind, and be
drawn all to the same scale.
The process of addition consists in drawing the lines in
position, end to end, all in one direction, so as to form a
straight line equal to the sum. If lines are to be sub-
tracted, they must be set off in the opposite direction.
A' ' o-A-->* B {
\ 1 ! k C 4
Cl , j '^._ D ._._^
D\ 1 1 ! 1 Ljy
PR S Q
Thus in the figure OP= A ; OQ=A+B; OR = A +
B- C; OS=A + B- C + D
Note 1. Observe that the sum OS would be the same if the 'terms
were added in any other order, say in the order D + B + A C.
Note 2. The summation is most readily effected by applying the
edge of a strip of paper in succession to the lines A, B, C, D, and
marking the points O, P, Q, R, S with a sharp pencil.
377. Multiplication and division. The product of two
lines, like that of two numbers, means that one is to be re-
peated as often as there are units in the other. That is
Unit line : one line : : other line : product.
And in division we have the proportion
Divisor : dividend : : unit : quotient.
Thus graphic arithmetic consists mainly of finding fourth propor-
tionals to given lines. The results may be represented graphically, or
expressed numerically by measuring on the unit scale.
4 S8 GRAPHICS chap.
378. Problem. Having given two lines A, B, and the
unit line S ( = V), to find the product of A and B.
Let AxB=X=Xx i=XxS.
A X
From which =, =tt5 ox S : A : : B : X.
o h
We thus find X as a fourth proportional to .S', A, B as follows.
Draw two lines intetsecting at O.
Along one set off OA equal to A. Along the other set
off OB and OS equal to B and S.
Join SA, and draw BX parallel to SA.
Then OX represents the required product graphically.
Measuring OX on the scale of S, or -h", to one unit, the
product is 2.94.
379. Problem. Having given two lines A, B, and the
line S ( = J"); to find the quotient A -=- B.
Let 4- = X = -= ~ . Or B : A : : S : X.
B 1 6
So X is a fourth proportional to B, A, S, and may be found thus :
Draw two lines intersecting at O.
Along one set off OA equal to A. Along the other set
off OB and OS equal to B and S.
Join BA, and draw SX parallel to BA.
Then OX represents the required quotient graphically.
Measuring OX on the scale of .S", or |", to one unit, the
quotient is 2.25.
Note. With a different unit A and B would represent other numbers,
but their quotient when measured would be the same number.
380. Problem. Having given two lines A, B, and their
product X, to find the unit line S.
This problem is the converse of Prob. 377-
In Fig. 378 set off along one axis OX and OA equal to
X and A. Along the other set off OB equal to B.
Join XB, and draw AS parallel to XB.
Then OS is the required unit length.
381. Problem. Having given two lines A and B, and
X the quotient A-fB, to find the unit line S.
This problem is the converse of Prob. 378.
XX
GRAPHIC ARITHMETIC
489
o
378
.K
A
A
B
1 II 1
X
I
A
B
5\
n 1
379
X
A
In Fig. 379 set off along one axis OX and OA equal to
X and A. Along the other set off OB equal to B.
Join ^4i?, and draw XS parallel to AB.
Then OS is the length of the required unit.
Examples. 1. The unit line being 1.34" long, construct the unit
scale, and mark off a length representing 2.27 units.
2. If the line S, Fig. 379, represent unity, what numbers do the
lines A and B represent? Ans. 3.60 ; 1.62.
3. In Fig. 379, taking S as the unit of length, find the number of
units of area in the rectangle, two adjacent sides of which are
equal to A and B. Ans. 5.83.
4. Find the line which represents |- to a unit of -J-".
5. In Fig. 379, if S be the unit, represent graphically :: .
A
6. Draw the lines A and B of Fig. 379 double the length. Then
A - B
determine the numerical value of r rv Ans, o. ?86.
A + B
7. In Fig. 379, if 5 represents the product of A and B, find and
measure the unit of length. Ans. 2.92".
8. If the line ./, Fig. 379, represent the fraction , determine
3-9
and measure the unit of length. Ans. 1.02".
490 GRAPHICS chap.
382. Problem. Having given the lines A, B, C, . . .
and the unit line S ( = |"), to find the continued product
A x B x C x . . .
The construction is a repetition of that of Prob. 37S.
Draw two lines intersecting at O.
Along one set off OA equal to A. Along the other set
off OS, OB, OC, . . . equal to S, B, C, . . .
Join SA, and draw BX 1 parallel to SA. Join SX V and
draw CX, parallel to SX 1 ; and so on.
Then OX x = Ax=i.s; OX 2 = A x B x C= 2.6, etc.
383. Having given the lines A, B, C, . . . and the
unit line S ( = I"), to find the value of - -
B x C x . . .
The construction is a repetition of that of Prob. 379.
Along one axis set off OA equal to A. Along the other
set off OS, OB, OC, . . . equal to S, B, C, . . .
Join AB, and draw SX 1 parallel to AB. Join CX V and
draw SX, parallel to CX t ; and so on.
A A
Then OX 1 = = 1.93 ; OX, = -g^-= I * 5 ' et '
B D
384. To find the value ofAx-x-x . . .
Set off the lengths A, B, C, D, E, . . . from O along
the two axes in the manner shown in the figure.
Join CB, and draw AX X parallel to CB. Join ED, and
draw X X X^ parallel to ED. And so on.
Then OX, = A x . OX, = Ax x , etc.
1 C C E
Note. The Roman numerals in the figures indicate the order in
which the dotted lines between the axes are drawn. A reference such
as II. || I. means that line II. is to be drawn parallel to line I.
Examples. 1. In Fig. 383, if S be the unit of length, find the
number of units of volume in the rectangular prism of which A,
B, and Care three edges. Ans. 19.8.
2. Determine graphically the value of 3.14 x i.7 2 ^o.67.
Ans. 13.54.
C E . ...
3. In Fig. 384 determine A x -^ x , the unit being i . Ans. 7.3.
XX
GRAPHIC ARITHMETIC
491
J I || I
X z X,
383
A
384
Mr
C D X 9
-A
-If
JL II T
w II m
492 GRAPHICS chaf.
A C
385. Problem. To find the value of + - + . . .
B I)
The method of solution is to reduce the fractions to a common
denominator K=k units, where k is any known number (say 3).
A A' C A'.,
Thus let - B = -^; ^ = ^; . . .
A C
Then X t = Xx -= ; A 2 =A'x-vr; . . . and these values may be
found as in Prob. 383.
Set off OK= K, say f ", OA A, . . . along any con-
venient axes in the manner shown.
Determine X v A'. ... by drawing the pairs of parallel
lines between the axes in the order indicated by the Roman
numerals.
A C
Then + + . . . = OX, + OXc, + . . . measured on
Jy JJ
the unit scale and divided by k ; or measured directly on
the scale on which OK (f") is the unit.
386. Problem. To find the value of A\B + CD + . . .
The series of products are first reduced to any convenient common
base A' (|") representing k units, k being a known number (say 3).
Thus let A-B = X 1 -JsTj C-D=X^K; ....
Tl V A ' B V C ' D
Then X 1 = ; X t = -g- ; . . .
These values may be found as in Prob. 383.
Make the construction which is sufficiently indicated by
the notation of the figure.
Then AS + C-D + . . . = OX x + OX 2 + . . . measured
on the unit scale (\") and multiplied by k (3) ; or measured
directly on the scale on which OK (f ") = U 2 (or 9) units.
387. Probl*em. To find the number of units of area in
a given irregular polygon, having given the unit of length
S.
First reduce the given polygon by Prob. 30 to an equivalent tri-
angle, then find half the product of the base and altitude of the triangle.
In the figure OA = altitude ; OB~\ base ; OS= unit length. Then
OX measured on the unit scale gives the required area.
XX
GRAPHIC ARITHMETIC
493
X, X ? A
w
a
C AX
R II I
W || IE
X,
A
B
< C
J?
387 b
Hi
r bo
A X.
494 GRAPHICS chap.
388. Involution and evolution. Involution is the pro-
cess of raising a number to any integral power. Thus A n
(read A to the n\h power) stands for the continued product
A x A x . . . , where the factor A occurs n times. n is
called the index of the power, or simply the index. The
value of A n can be found by Prob. 381, putting B, C, . . .
each equal to A. It may also be found by the general con-
struction of Prob. 394. In this case, however, the special
method given in the next problem is preferable.
Evolution, or the extraction of roots, is the inverse of
involution ; thus v A, the nth. root of A, denotes a quantity
which being raised to the nth. power gives A as the result.
The general graphical solution requires the use of a logarith-
mic curve or spiral, and is given in Prob. 394. The extrac-
tion of the square root, and of the fourth, eighth, sixteenth
. . . roots may however be effected by simple geometry.
See Probs. 390 and 391.
389. Problem. Having given a line A, and the unit
line S, to find the integral powers of A, viz. A' 2 , A 3 , .
Also to find the integral powers of the reciprocal of A, viz.
1-7- A, 1-A 2 , . . .
Draw two perpendicular axes intersecting at O.
Along them set off OA and OS equal to A and S.
Join SA, and draw the lines AX, XX V . . . SY, YY V
. . . alternately perpendicular and parallel to SA, as shown
in the figure.
Then OX=A 2 ; OX l = A 3 ; . .
And OY=~; OY.=~; . . .
A' A %
These results follow from the similarity of all the tri-
angles formed by the axes and the dotted lines.
Note. If OS had been made equal to s units on the scale for A,
then the scale for measuring OX ox A' 1 would have been OS to s' 2 units;
and for OX 1 or A 3 , OS to s 5 units ; . . . And the scales for measuring
OY. OY,, . . . i.e. , , r , . . . would have been OS to - , -, . . .
1 A A* s s z
unit respectively.
XX
( ; RAP] IIC ARITHMETIC
495
X_LI
111
4
1/
v.
x,
Y/
o X>4
389
\
nrN
r
Y
l
fa
f
/
V
t
/
/
X
Examples on Problems 384 to 388.
Q M
1. Find the value of -", + -54--^. Ans. 4. 86.
2. Find the value of P'Q+Q-R + M'JV, the unit being 1".
Ans. 1 5. 8.
N
3. Find the value of + Q'M, the unit being R. Ans, 5.49.
4. If P be the unit of length, find the values of Q 2 , Q :i , , , , --.
Ans. 5.35, 12.37, .432, .187, .0808.
5. Find the line which represents the cube of a line 1.37" long to a
unit of 1". Ans. 2.57".
6. If the line N represent the square of the number represented by
the line R, find the length of the unit. Ans. 1.3".
7. A line 2.9" long represents the sum of the areas of an equilateral
triangle and square, each of 1" side. Determine the unit of
length. Ans. .494".
8. Draw a curve which will give the squares of all quantities from o
to 6, taking J" as the unit.
496 GRAPHICS chap.
390. Problem. To find the square root of a given line
A, having given the unit line S.
Let A = J A ; then A" 2 = A=Axi=AxS.
That is -rr= ''; A : X : : X : S.
Thus the required square root X is a mean proportional between A
and S. See Prob. 27.
Draw a straight line, and from any point O in it mark off
in opposite directions OA and OS equal to A and .5'.
Bisect SA in C, and with centre C draw the semicircle on
SA. Draw OX perpendicular to SA.
Then <9X ( - 1.54) = sj OA x OS= J~aVi= J At.
If S be not a unit line, OX represents the square root of A x S.
Note 1. If the construction were repeated on OX, the 4th root
OA would be obtained. And by continued repetition the 8th, 1 6th,
. . . , 2th roots might be found. See next problem.
Note 2. If A were a large or small quantity compared with the
unit .S", an "ill-conditioned" construction might be avoided by making
OS equal to s units, where s is any convenient known number. In
which case the square root OX would require to be measured on the
scale of OS to *Js units ; the 4th root on the scale of OS to *J s units ;
and so on. See next problem.
391. Problem. To find the eighth root of 614.
To any convenient scale set off OA equal to 614 units,
and OS equal to say 2 s , that is 256 units.
Draw OX perpendicular to SA, and describe the semi-
circles in the order figured, making OB, OC equal to OX,
OY.
Then ^614= OZ measured on the scale on which OS
= v 256 = 2 units.
392. Problem. To draw a figure which shall give the
square roots of the first n natural numbers.
Draw OA of unit length on any convenient scale.
Draw AB perpendicular and equal to OA. Join OB.
Draw BC perpendicular to OB and equal to OA. Join
OC. And so on.
Then OB= J 2; OC= Jz; . . . (Euc. I. 47-)
XX
GRAPHIC ARITHMETIC
497
s o c
A
S
390
OB =
CJC
, OC = OY
X
-__
, '
Y
/ *
z~-
/
1 _
! ': i
S O CB A.
391
392
Examples. 1. Determine graphically \/s.2, ^273, and
\l 0.035. .4ns. 2.28; 16.5; .187.
2. Determine ^165, v/o.165, .3I4 3 > and .031 4 3 . Am. 3.58;
.637 ; .031 ; .000031.
3. A line 2.6" long represents 4^ units. Obtain lines which repre-
sent 1. 3, V, and 2 ^"6 units. Am. .75"; 1.93"; 2.83".
4. On page 495, taking P as the unit line, determine the line which
represents JiV-^Q. Ans. .694".
5. If the line N on page 495 represent the square root of the line
Q, determine the unit line. Ans. 6.06".
6. If a line i\" long represent J 2, determine a line representing
J J. Ans. 2.16". 5
7. Taking V as unit, obtain lines representing 15, Ji$, ^
Ans. 3-75"; -97"; .323"- VIS -
8. Taking the lines/ 5 and Q on page 495, determine JQ^ + P 2 ,
JQ 2 - P\ and the ratio sj ? ~ ]. Unit = o. 5".
Hint. Make use of Euclid I. 47. Ans. 4.03; 3.33; 1.21.
2 K
498 GRAPHICS chap.
393. Examples worked out.
1. in and n are two given lines, determine s/mln, the unit bsing
0.25".
Set off OS, Fig. 1 (a), equal to the unit, 0.25".
In the opposite direction set off OFF, M equal to tn and ;/,
and describe semicircles on SN and S3I as diameters.
Draw 0M X at right angles to OM, then 0M X is dm, and
07V X is \fn.
Set off these lengths and OS ( = j") along two axes on (/>) as
shown. Join I\I X N X , and draw SX parallel to M X N X .
Then OX represents \'m/n, and measured on the j" scale
is 1.5.
s/t'AB-CD
2. Find a line which represents p= . Unit=\\ . (1896)
_ v 7 3 #
sJ^-AB-CD^r V3 is the same as \UaB-CD. Set off
AB and divide it into three equal parts as shown in Fig. 2 :
make BE equal to one of the parts ; then AE \AB.
Make FF= CD ; on AF describe a semicircle, and erect
the perpendicular EG ; then EG represents sj%CD'AB.
3. If ^^.^R, find the unit of length. Fig. 3.
This may be written
Px Q R . R 2V
~v Ar~ ~* ' or ' unlt ~-^ x ~r> x ~r\'
R x JV unit P Q
We may now proceed as in Problem 384.
. A' 2 N
Or, write, unit = = . y
Then set off OP=P; and, at right angles, OR = R.
Toin PR, and draw RA" l perpendicular to PR to meet PO.
Then 0X X represents R 2 +P; for OP-OX l = OR 2 .
Make OQ and OJV equal to Q and jY ; join QN, and draw
X X X 2 parallel to QN.
R 2 A r .
Then 0X 2 represents , or unit line; it measures 0.76".
4. Determine the value of (o. 2S9) 3 . Fig. 4.
Set off OA equal to .289 on the scale ' to o. 1.
Set off OS equal to V', i.e. . 2 or i unit.
Join AS, and draw AX, XX X respectively perpendicular and
parallel to AS. Then 0X X represents (.289)".
Measuring 0X X on the scale of OS to (.2) 3 of a unit, that
is, V' to .008 unit, or -^" to .001 unit, the answer is .024.
XX
GRAPHIC ARITHMETIC
499
17t\-
/!
/
Rgl
/
(M,
^~~~
/
s*
1 /
1/
N,
1/
1/
1
1
(a)
K/ (b)
S/\
S
If
\ v.
_-- G
L
[\
\
v
\
MO X M,
Ai
c
B
J)
Q.
B E
Fig. 2
F
s;
p
O NX,
^T^ 5
<R
~3.
'/ y
Fig.,
R
Q
N
A
O
1
x,
/ Fig. 4.
500 GRAPHICS chap.
394. Problem. Having given a line A and the unit
line S, to find the value of A", where n is any given
integer or fraction, positive or negative.
By the construction of Problem 129 set out on the base
BB any suitable logarithmic curve LL.
Determine OS, the ordinate of unit length S.
On OS set off OA equal to A. Draw AP, PM par-
allel and perpendicular to BB.
Mark off ON = n times OM; in the same direction as
OM if n is positive, and opposite if n is negative.
Draw JVQ, QX perpendicular and parallel to BB.
Then OX measured on the scale of unit S gives A n .
Note 1. The logarithmic curve might be used to perform graphic
arithmetic in a manner exactly analogous to the way in which a table
of logarithms "is employed in making arithmetical calculations.
Ordinates such as PM, QN, represent numbers, and the abscissae
OM, ON, their logarithms.
Note 2. The logarithmic curve might be replaced by a logarithmic
spiral, see Prob. 128.
395. Miscellaneous Examples.
1. Determine graphically 2.5s, i.9~?, and 0.72 -0 ' 6- Ans. 1.84 ;
.76 ; 1.22.
*2. If -^=C, find the unit of length. Ans. if". (1898)
*3. Find a line which represents \t ^'A'B-r- V3.
(1896)
*4. Copy the figure double size. Then reduce it to an equivalent
triangle with its base on All and its vertex at E ; and find the
length of a line representing the area of the figure, taking for
unit 1". O894)
c /a*-b 2
*5. Find the ratio , \J ^ 2 , the unit of length being 2".
Ans. .932". Hint. Compare Ex. 8, p. 497. (1890)
6. Determine by graphic arithmetic a line representing the con-
tents of a rectangular solid whose dimensions are 3" x 1.75" x
1.25". Unit = 2.5". Ans. 1.05" long. (1897)
7. Find a line to represent x 5 when x = 2 \J 1 - \ 3. Unit= 1".
Ans. 1.32". (1895)
8. A line 3" long represents the sum of the areas of a pentagon,
square, and equilateral triangle of 1" side. Determine the unit
of length. Ans. 1.025". O891)
XX
GRAPHIC ARITHMETIC
5oi
N B
C\
D
^B
As
3 B>
CHAPTER XXI
GRAPHIC STATICS
396. Directed quantities. In arithmetic and algebra we
are concerned with number and magnitude. A second
important quality possessed by many objects and phe-
nomena, and which can be made the subject of calculation,
is that of having a definite direction in space. Graphical
processes are uniquely fitted to deal with this property.
Every one is acquainted with such quantities, and would,
for instance, distinguish between a length measured to the
north, and the same measured, say, to the east.
The student will readily call to mind other instances in
which direction is associated with magnitude. He may
think of a force of so many pounds weight, and acting along
some definite line ; of a top spinning at a definite speed
about an axis, perhaps inclined ; or of a field of force,
which at any point has intensity and direction. The word
vector is a general term, used to denote all such directed
quantities.
Some kinds of quantities like volume, time, mass,
temperature, energy, are essentially non-directive ; others,
like forces, velocities, which have direction, may yet be
treated arithmetically as to magnitude only. But in the
class of problems we are now about to investigate, it is a
cardinal feature that the directions of the quantities which
enter into the case, as well as their magnitudes, shall have
influence on the result.
CHAP. XXI
GRAPHIC STATICS
503
J
1
L
a
Scale I " to 50 feet
397. Graphic representation of a vector. The relative
positions of points and their changes of position are amongst
the simplest examples of directed quantities. Such quantities
may evidently be represented with great convenience and
directness on paper by directed lines.
For example, a displacement of say 50 feet to the east
might be exhibited thus : Let the direction to the north be
indicated; then (1) draw a line running east and west; (2)
on this line mark off a segment representing 50 feet to a
suitable scale, and specify the scale ; and (3) to the segment
affix an arrow-head pointing eastwards. In the figure above,
oa measures 58 feet.
The change of position is thus represented completely
and without ambiguity, it being understood that the dis-
placement is a horizontal one. The arrow-head indicates
what is called the sense of the directed line ; it distinguishes
an eastward from a westward movement. Sometimes if is
convenient to indicate the sense in another way. The ends
of the line are marked and named in the order or sequence
corresponding to the beginning and end of the displacement.
Thus oa would indicate a displacement in the sense of to
a, or eastwards. If the ends of the line were named in the
sequence ao, this would be understood to indicate the
opposite or westward direction, from a towards 0.
Example. Mortlake is 2 miles west of Putney, and Wimbledon is
3 miles south of Putney. Plot the relative positions of these places to
a scale of 1 inch to the mile, and measure the distance and direction
of Wimbledon to Mortlake.
Aus. 3.6 miles, 33.7 west of north.
504 GRAPHICS chap.
398. Resultant and components. Rectangular com-
ponents. When a series of displacements are given to a
body, the change of position represented by the straight
line which joins the initial to the final position is called
the resultant displacement, and the several displacements
are called components.
Any displacement may be imagined to be made up of
components, and in an indefinite number of ways, just as
we may walk from one place to another by an indefinite
number of routes. But there is only one resultant which
corresponds to a given set of components.
We are said to compound a set of displacements when
we combine them to obtain their resultant, and in the
reverse operation we resolve a vector into components.
As just stated, a vector can be resolved into components
in an indefinite number of ways. For the case, however,
of two components which are parallel to given lines, there
is only one solution. If the two lines are perpendicular to
each other, the components are said to be rectangular.
Suppose a person to walk from one corner of a room
for a distance of 10 feet, in a direction making 30 with a
side wall. He would reach the same spot by walking
8.67 feet along the side, and then walking 5 feet parallel
to the end wall ; or he might first walk 5 feet along the end
wall, and then 8*67 feet parallel to the side wall. The
rectangular components of his change of position parallel to
the side and end walls are respectively 8.67 and 5 feet.
Thus the rectangular component of a vector in any direction
is obtained by projecting the vector on a line parallel to the
direction. This is often called simply the component in
that direction, rectangular being understood.
For example, the component of the given vector oa
parallel to the given direction XX, is equal to ob, or /;///,
obtained by drawing om, an perpendicular to XX, and ob
parallel to XX.
It is important that the student should be familiar with
the idea of resolving a vector in a given direction.
xxi GRAPHIC STATICS 505
X i > i X
m n
Examples on Articles 398 and 399.
1. A person journeys 1 mile eastwards and \ mile to the norlh-east,
find his resultant change of position. Find also the northerly
and easterly components of the resultant. Ans. 1.62 mile,
19. 1 N. of E. ; .53 mile, 1.53 mile.
2. A football is kicked a distance of 52 yards in a direction making
35 with the side lines of the field. Find the component dis-
placements of the ball forwards and sidewise respectively.
Ans. 42.6 yards, 29.6 yards.
3. Find the component of a vector 3.7 units long, in a direction
making 53. 6 with the direction of the vector. Ans. 2.18 miles.
4. The two rectangular components respectively southwards and
westwards of a horizontal vector are 31.3 feet and 20.7 feet.
Determine the magnitude and direction of the vector.
Ans. 37.1 feet, 56.5 S. of W.
5. A boy on a raft walks 9 feet from a point A to a point B, and
during this lime the raft moves (without rotation) a distance of
14 feet in a direction making 115 with AB. Find the re-
sultant motion of the boy. Ans. 1 3. 1 feet; 76. 4 with AB.
6. A ship at sea sails 8.7 miles through the water apparently to
the east, but an ocean current simultaneously carries the vessel
3.4 miles to the south-west. Find the resultant movement of
the ship, i.e. its displacement as regards the bottom of the sea.
Ans. 20. 9 S. of E. ; 5.88 miles.
7. A body receives component displacements of 7.38 and 5.16 feet
in two directions which make 7 J -3 with one another. Find the
resultant displacement. Ans. 9.58 feet, 47 with second vector.
8. Two places are 1. 29 miles apart in the direction north and south.
A person journeying from one to the other first walks to the
north- north-east, and then to the north-west. Find the lengths
of the two stages of the journey. Ans. .987 mile, .534 mile.
9. If a person walk between the two places of the preceding
example in two straight paths, one of 1 mile, the other of -^
mile, find the directions of the paths. Ans. 20.6", 44.8".
!
506 GRAPHICS chap.
399. The triangle and parallelogram vectors. Suppose
an object to be placed on a table in a room. Let the
position of the object be changed on the table in the
manner represented by the directed line P, and let the
position of the table in the room be also changed as re-
presented by Q. (The table must remain parallel to itself.)
It is required to find the resultant displacement of the
object in the room.
From any point a draw ab equal and parallel to P and
similarly directed ; and from b draw be equal and parallel
to Q and similarly directed ; mark ac with an arrow-head
pointing from a to c ; then the directed line ac represents
the resultant displacement. The triangle abc with the
arrow-heads or their equivalent is called the triangle of
displacements. It is a vector triangle.
If the movement of the object on the table had pre-
ceded the movement of the table, the path of the object in
the room would have been similar to ab, be ; if the displace-
ment of the table had been effected first, the path would
have been similar to ad, dc, where abed is a parallelo-
gram ; if the two movements had occurred simultaneously,
each at a constant rate, beginning and ending together, the
path would have been a straight line similar to ac. Any
path, stepped or curved, could be effected by suitably
timing and adjusting the two component displacements ;
but in all cases, when the component changes directed by
/'and Q are fully effected, the resultant is the same, viz. ac.
In the problems which follow, we are seldom concerned
with the actual paths.
Let A be the initial position of the object in the room ;
draw AC parallel and equal to ac, and similarly directed,
then C is the final position of the object.
The law of the triangle is of fundamental importance ;
two vector quantities of any like kind are compounded by
the same rule. The student should pay great attention
to the rule. Observe that, for compounding P and Q, two
triangles, which are similar, may be drawn ; that the two
XXI
(IRAPHIC STATICS
507
p's are similarly directed in the triangle, and also the two
^'s ; that these directions are with the motion of a point
which travels round either triangle against the resultant ac.
This last property may be expressed by saying that in either
triangle the components p and q are circuital, and the
resultant r is non-circuital (with/ and q).
Note also that in the parallelogram abed the three
vectors p, q, r which meet at a are all directed away from
a, and those that meet at c are all directed towards c.
The resultant of P and Q might therefore be found as
the diagonal r of a parallelogram of which p and q are
adjacent sides ; />, q and r being directed all towards, or
all away from, the common point where the three meet.
This construction is known as the parallelogram of displace-
ments ; it is exactly equivalent to the triangle of displace-
ments.
We thus have the following equivalent rules for obtaining
the resultant of two given vectors :
Rule 1.- Place the two given vector lines end to end
circuitallv, and obtain the non-circuital closing side of the
vector triangle.
Rule 2. Obtain the diagonal of a parallelogram which
//as the two given vector lines as adjacent sides, the three lines
being all directed away from, or toivards, their common end
point.
508 GRAPHICS chap.
400. The vector polygon. Let P, Q, P, S represent
any component displacements which a body receives ; it is
required to find the resultant displacement.
Starting from any point a, Fig. ( i ), draw ab, be, cd, de
respectively, equal, parallel, and similarly directed to P, Q,
P, S; join the first point a to the last point e, and direct
it from a to e ; then ae represents the resultant.
For, by the rule of the triangle, ac is the resultant of ab
and be ; ad is the resultant of ac end cd, and therefore of
ab, be, cd; and ae is the resultant of ad and de, and there-
fore of ab, be, cd, de. The vector polygon abede is called
the polygon of displacements, and ae is its clositig side.
In drawing the polygon the components may be joined
in any sequence it will be found that in all the polygons
which are thus possible the closing sides will be vectorally
equal (i.e. equal, parallel, and similarly directed). Fig. (2)
shows the polygon when the sequence is S, Q, P, P;
the closing side of (2) is seen to give the same resultant as
the closing side of (1).
Observe that in both polygons the component sides
p, q, r, s are directed circuital!) 7 , and the resultant side
v non-circuitally with the others.
The polygon (3) has been drawn to show the student that
if the component sides are not all circuital, the answer is
incorrect ; the closing side does not then give the resultant.
The vector polygon may be applied to find the resultant
of any number of components.
If the end point e had coincided with the starting-point
a the resultant would have been zero, and all the sides would
have been circuital. Thus a body which receives com-
ponent displacements, represented by the circuital sides of
any closed polygon, has no resultant displacement.
The rule for finding the resultant of a given system of
vectors may be stated as follows :
Rule. Place the given vector lines end to end circuit-
ally, and obtain the non-circuital closing side of the vector
polygon.
XXI
GRAPHIC STATICS
50Q
Examples. 1. A body receives component displacements of 2,
5, and 7 units in directions parallel respectively to the sides of
an equilateral triangle taken circuitally. Find the resultant dis-
placement. Ans. 4.3. 83.4, 36.6 with 3 and 5.
2. Draw a quadrilateral ABCD having given AB = 3.J, BC2..'],
CD =5.2, 75^ = 3.4, AC=^.i t . Suppose a body to be moved
4 feet parallel respectively to each of the directions AB, BC,
CD, AD, AC ; find the resultant movement. Show, by draw-
ing a figure, that if the displacements had taken place in any
other sequence, say the sequence BC, AD, AB, AC, and CD,
the resultant displacement would have been the same both in
magnitude and direction. Ans. 11.4 feet, 6o with AB.
3. A person walks 29 steps east, 51 steps south-south-east, and 13
steps south-west. Find how many steps respectively west and
north will bring him back to his original position.
Ans. 39.3; 56.3.
4. A body is moved 10 feet parallel to a line OX, and 15, 20, and
25 feet each in directions making respectively to 30, 90 , and
'35 with OX. Find the component displacements parallel and
perpendicular to OX, and the resultant displacement.
Ans. 5.31 feet, 45.2 feet, 45.5 feet, 83. 3 .
510 GRAPHICS chap.
401. Concurrent systems of forces. All vectors have
magnitude, direction, and sense. Some, like linear veloci-
ties and couples, have no particular location in space ; they
are completely represented by a directed segment of a line
which is free to be moved parallel to itself either laterally
or longitudinally or in both ways. Others, such as rotations
and forces, take place along definite lines ; they are repre-
sented by a directed segment of the line ; the segment may
have any position in the line, but may not be moved
laterally out of the line.
In the problems which follow, the vectors will be con-
fined to forces. To specify a force we must define its
magnitude, direction, sense, and any one point in its line of
action.
When there are several forces acting on a body, we
speak of them collectively as a system of forces. If the
lines of action of all the forces pass through a common-
point, we have a concurrent system. In the first instance
we shall direct attention to such systems.
The general propositions on unlocalised vectors already
given apply to forces, with the restriction that forces are
vectors localised in lines.
Thus to compound a system of concurrent forces, we first
construct a' polygon of forces, that is a vector polygon drawn
as if the forces were unlocalised. The resultant is then
represented by a line drawn through the common point of
the system, and equal, parallel, and similarly directed to
the non-circuital closing side of the force polygon. If the
force polygon close, i.e. if the first and last points coincide,
the resultant is zero. In this case the system is in
equilibrium, the components balancing one another.
A force which being added to a system of forces pro-
duces equilibrium is called the equilibrant of the system.
It is equal in magnitude to the resultant, and acts in the
opposite direction along the same line.
We now state some theorems relating to concurrent
forces.
xxi GRAPHIC STATICS 511
Theorem 1. The resultant of a concurrent system of
forces is represented by a line through the common point,
equal, parallel, and similarly directed to the non-circuital
closing side of a force polygon.
Theorem 2. // is necessary and sufficient for the equili-
brium of a concurrent system of forces that a force polygon
shall close.
Theorem 3. Two forces which are equal in magnitude
and which act in opposite directions along the same line
balance one another.
Theorem 4. Two forces which are in equilibrium must
be equal in magnitude, and must act in opposite directions
along the same line.
Theorem 5. The resultant of two forces passes through
their intersection, and its magnitude, direction, and sense may
be obtained by constructing the triangle or the parallelogram
of forces as described in Rules 1 and 2, Art. 399.
Theorem 6. Three concurrent forces will be in equilibrium
if their magnitudes and senses can be represented by the
circuital sides of a triangle of forces.
Theorem 7. /// order that three forces mar be in equili-
brium they must be concurrent {unless they are parallel), and
their magnitudes and senses must be represented by the
circuital sides of the triangle of forces.
Theorem 8. If forces which balance one another be added
to or subtracted from any system of forces, the resultant of the
system is unaltered.
Theorem 9. If a system of forces is in equilibrium, any
force reversed in sense is the resultant of the others.
Theorem 10. The algebraical sum of the components of a
system of forces in any direction is equal to the component of
the resultant in the same direction.
These theorems are collected for easy reference, but the
beginner will only gradually come to understand them as he
applies them to problems. He should consult them from
time to time as he works the succeeding examples of this
chapter.
512 GRAPHICS chap.
402. Problem. Forces of 5 and 2 units act respectively
to the right and upwards along the given lines OM, ON
which include an angle of 70 . Required the resultant.
First Method. By the parallelogram of forces.
Let G be the intersection of the given lines.
To any convenient scale mark off from O, to the right
along OM, and upwards along ON, the lengths OP, OQ
of 5 and 2 units respectively.
Complete the parallelogram by drawing PR, QR parallel
to OQ, OP. Draw the diagonal OR, and direct it from
O to R.
Then OR represents the required resultant. It measures
6 units and makes 18.5" with OM and 51. 5 with ON.
Second Method. By the triangle of forces.
Draw any line ab parallel to OM, directed to the right
and 5 units long. From h draw be parallel to ON, directed
upwards, and 2 units long. Join ac, and direct it from a to
c, i.e. non-circuitally.
From O draw OR vectorally equal to ac, that is parallel,
equal, and similarly directed. Then OR represents the
resultant as before.
403. Problem. A given force is represented by OR.
Required its components along the given lines OM and ON.
Draw RQ and RP parallel to MO and NO. Direct
OP and OQ away from O, like OR is directed.
Then OP and OQ represent the required components.
404. Problem. A given force of 34 lbs. acts along OG.
Required its component parallel to LL, which makes 35
with OG.
Reciangtdar component is understood.
Select a convenient scale, say \" to 10 lbs., and mark
off OG 34 units long. Direct it from O to G.
Draw Off, GK parallel to LL, and OK, GLL perpen-
dicular to LL. Direct Off and OR like OG, away from O.
Then Off represents the required component parallel to
LL, and measures 27.9 lbs. The perpendicular component
O K measures 19.5 lbs.
XXI
GRAPHIC STATICS
513
405. Problem. The rectangular components of a force
are 7 and 5 lbs. Find the force.
Draw any two perpendicular lines, along which, to a
convenient scale, mark off from their intersection O, OA
and OB to represent 7 and 5 lbs.
Complete the rectangle contained by OA, OB, and draw
the diagonal OC. Then OC represents the required force ;
it measures 8.60 lbs. and makes 35.5 with OA and 54.5
with OB.
406. Problem. A small ring is at rest under pulls of
2, 3, and 4 lbs. exerted through strings. Find the angles
between the strings.
Draw a triangle of forces def with, its sides 2, 3, and 4
units long and circuitally directed.
From the centre of the ring draw lines vectorally equal to
the sides of the triangle.
These represent the three pulls, and the angles D, E, F
between them are found to measure 133.4 , 7 5. 6, and 15 i.
2 L
514 GRAPHICS chap.
407. Problem. A load of 2 tons is suspended from
a pin A, which is maintained in position by the pull of a
horizontal tie AL, and the thrust of a strut MA inclined
at 25. Find the pull and thrust.
Draw ab 2 units long and directed vertically downwards
to represent the given load. From the ends a and b draw
two lines parallel to the tie and strut to intersect in c, and
direct the sides be, ca circuitally with ab.
Then abc is the triangle of forces representing the equili-
brium of the pin A. The pull of the tie is found by
measuring ca to be 4.28 tons. And the thrust of the strut
AT A is 4.72 tons, as given by be.
408. Problem. Four given pulls and thrusts K, L, M,
N act on a point as denned in the figure. Find their
equilibrant.
Construct a vector polygon thus : Select a convenient
scale (say |" to 10 units), and from any point a draw ab,
be, cd, de respectively parallel and similarly directed to K,
L, M,N, and 72, 40, 105, and 58 units long.
Join ea and direct it circuitally, i.e. in from e to a. A
line R drawn through the common point vectorally equal to
ea represents the required balancing force.
Measuring ea, the magnitude of R is found to be 88.5
units. The angles E and A measure 45 and 121.
Observe the two systems of notation. In one, the forces are
denoted by A*, L, M, N, R, and the corresponding sides of the force
polygon by k, I, m, n, r.
In the other, known as Bow's notation, the angles round the
point are lettered A, B, C, D, E, and the corresponding corners of the
force polygon a, b, c, d, e. In this system a force is referred to by
naming the two letters in the angular spaces on each side of it. Thus
the force " BC" (or CB) would mean the force "Z," and the corre-
sponding letters b, c would appear at the ends of the side /.
On going round the point (clockwise), the sequence .-/, B, C, D, E
of the angles is seen to be the same as the circuital sequence a, b, c, d,
e of the corners of the vector polygon. If we agree that the two letters
which denote any force shall be named in the sequence derived from
the rotation, e.g. BC for L, then the same sequence, be, agrees with the
circuital arrow. This convention will be used later.
XXI
GRAPHIC STATICS
515
Tons
L\40^76--M=I05
K=7Z a \ J7
^ V N-58
R>
\
Examples. 1. Find the force whose horizontal and vertical com-
ponents are 72.2 lbs. and 45.6 lbs. Ans. S5.3 lbs. inclined at
32. 3-
2. A force of 100 lbs. acts at an angle of 53. 7 with the horizontal :
find its horizontal and vertical components. Ans. 59.2 lbs ;
80.6 lbs.
3. Two forces of 3.5 and 6.7 lbs. act in directions which include an
angle of 75; find their resultant. Ans. 8.32 lbs., making 24
with the larger force.
4. The lines of action of two forces include an angle of 120 ; one
of the forces is 10 lbs., and their resultant is 9 lbs. Find the
other force. Ans. 7.4 or 2.62 lbs.
5. A horizontal string 8 feet long is
and from a point 3 feet from one
end a load is hung which is in-
creased until at 10 lbs. the string
breaks, the deflection when this
happens being 6". If a piece of
the same string were to hang
vertically, what load would it
carry? Ans. 37.9 lbs.
6. Determine the resultant of the con-
current system of forces defined by
the figure. Ans. A pull of 54.4
\7Z
f/43\
128 \*
making 20. 4' with the 41 force, and 17. 6 with the 63 force.
two walls,
9Z
34
516 GRAPHICS chap.
409. Problem. Forces which act at a point are in
equilibrium ; all except two, M and N, are known com-
pletely, but only the lines of action of these latter are
given. To find the magnitudes and senses of M and N.
The problem is solved by applying Theorem 2, Art. 401.
We shall again illustrate Bow's notation, and to be able to apply
this the figure has been first prepared by drawing the lines of action
of the forces each with one end at the common point, so that there
shall be the same number of angular spaces as there are forces. It has
also been arranged that M and N shall be adjacent so as to bound one
of the spaces ; this is always possible, since a force may be represented
either as a pull from one side or as an equal push from the other.
The spaces are then labelled A, B, C, D, X. Clockwise rotation
being adopted, the construction may be described as follows :
Draw ab, be, cd the three sides of the force polygon cor-
responding to the three given forces AB, BC, CD.
Close the polygon by drawing the two remaining sides
dx, ax parallel to DX, AX, intersecting in x.
The required forces M and JVare then given by the sides
dx, xa taken circuitally, or as determined by the clockwise
sequence DX, XA. M is seen to be a pull, and N a thrust.
410. Problem. A load W is suspended from a jib
crane ; to find the thrust in the jib CA, and the pull in
the tie AB.
First consider the equilibrium of the tie. At the end A it is in con-
tact with the pin, and between the two surfaces there is a force action
consisting of equal and opposite forces, one on the pin, the other on
the bar. If the joint is supposed frictionless, the line of action of these
forces must pass through the centre of the pin.
A similar force occurs at the other end B of the tie. Neglecting
its weight, these are the only two forces acting on the bar, since it is
nowhere else in contact with anything. And as the tw<o forces balance
they must be equal and opposite. The bar must therefore be subject
to a direct pull or a direct thrust. The tie is shown separately under
forces T, T indicating a pull.
So for the jib. And generally, in a hinged frame, loaded at the
Joints only, the forces at the ends of any bar must be equal and opposite,
and must act along the line joining the centres of the pins.
To obtain these forces, and to distinguish pulls from thrusts, we
must draw the closed polygons of forces for the pins which hold the
frame together at the joints.
XXI
GRAPHIC STATICS
5i7
409
Consider the forces on the pin A. There is the vertical force JV,
being the load suspended from it. There is the force which the tie
exerts on it acting in the line of the tie. And there is the force exerted
by the jib along the line of the jib. The pin is shown separately with
the lines of action of these three forces, the angular spaces round the
pin being labelled A', Y, Z.
To draw the triangle of forces for tlie pin A. Adopting clockwise
rotation, the sequence of letters for W, the known force, is X Y. So
set off xy vertically downwards to represent //". Draw xs, yz parallel
to XZ, YZ. By watch - hand rotation about the pin obtain the
sequences YZ, ZX for jib and tie, and direct those lines near the pin
A to agree with the sequences yz, zx of the triangle of forces.
YZ or .S" is directed towards the pin A, indicating a thrust in the
jib ; and ZX or T is directed away from the pin, denoting a///// in the
tie. The magnitudes are obtained by measuring yz and zx.
518 GRAPHICS chap.
411. Problem. A hinged triangular frame is in equili-
brium under given forces P, Q, R applied at its corners.
To find the forces in the three bars of the frame.
Since the forces P, Q, A' maintain equilibrium, their lines must all
pass through one point (not shown), and their magnitudes and senses
must be given by the circuital sides of the triangle of forces abc.
To employ Bow's notation, the lines of the applied forces P, Q, R
are all drawn outside the frame, and with their ends at the joints. The
external and internal spaces are then lettered A, B, C, D.
To obtain the forces in the bars, we may draw the triangles of forces
for the three joint pins. Clockwise rotation is adopted.
The tipper joint. The spaces round this joint named in watch-hand
sequence are A, B, D. The triangle of forces named circuitally is
abd. The forces BD, DA, directed in the senses bd, da, both act to-
wards the joint, and indicate thrusts in the two inclined bars.
The left-hand joint. The spaces in clockwise sequence are C, A,
D. The triangle of forces is cad. The forces AD, DC, directed as
ad, dc, act the first towards, the second away from the joint. The
horizontal bar of the frame is thus in tension.
The right-hand joint. The spaces are B, C, D, and the triangle of
forces is bed.
The force diagram. The four triangles of forces may be superposed
on one another so as to form the figure abed. The six lines of the
latter measured to scale give the magnitudes of P, Q, R and of the
forces in the bars. It is called the force diagram for the frame. The
arrow-heads would conflict and are omitted, but the sense of any force
may be obtained by the convention of Bow's notation, rotation about
the joints being always clockwise.
412. Problem. A string has its ends fixed, and four
forces P, Q, R, S act on it along the given lines. The
form taken by the string is shown in the figure. If P is
3 lbs., find the other forces Q, R, S, and the tensions in the
five segments of the string.
Bow's notation can be applied. The spaces are lettered A, B, C,
D, E, O. We shall again adopt clockwise rotation.
Draw ab 3 units long to represent P or AB. And draw ao, bo par-
allel to AO, BO to intersect in 0. Then abo is the triangle of forces
for the joint ABO.
Draw be, oc parallel to BC, OC, to intersect in c.
Draw cd, od parallel to CD,OD ; and de, oe parallel to DE, OE.
Then oabede is the force diagram. The forces P, Q, P, S can be
found by measuring ab, be, cd, de to a scale of -J" to 1 lb. ; and the ten-
XXI
GRAPHIC STATICS
519
a- d
Q 412
sions in the segments A, B, C, I), E of the string by measuring oa, ol>,
oc, od, oe.
Definition. The line of the string is called a funicular polygon or
link polygon. The point is its pole. The several lengths of the
string may be supposed to be replaced by weightless rigid links, con-
nected by hinged joints.
An indefinite number of such link polygons having for their pole,
or having new poles, could be found, all in equilibrium under the action
of the same forces P, Q, A\ S ; in some of these the links might be in
compression.
The link polygon plays an important part in graphic statics, as the
succeeding problem will show.
520 GRAPHICS chap.
413. Moments of forces. Couples. The constructions
hitherto Riven are not sufficient in themselves to solve com-
pletely the general problem on non-concurrent forces in one
plane. It may be shown that the magnitude, direction, and
sense of the resultant of such a system is given, exactly as
before, by the non-circuital closing side of the vector force
polygon ; but we still require to find the line in which the
resultant is located : this we can do by obtaining in addition
to the above any one point in its line of action.
Forces produce, or tend to produce, rotations in bodies
as well as motions of simple translation, and we now
require to measure such a tendency. This is done by find-
ing the moment as defined in the next paragraph.
Definition i. The moment of a force about any point is
the product of the magnitude of the force and the perpendicu-
lar distance from the point to the line of the force.
Thus in Fig. (<?) the moment of P about A'Px KM, where P\%
to be measured on the force scale, and KM on the linear scale.
It is seen that P tends to cause watctt-hand rotation about K. This
we may agree to consider as positive ; the opposite tendency is then
negative.
Definition 2. A pair of forces equal in magnitude, opposed
in sense, and acting in parallel lines, is called a couple.
Definition 3. The moment of a couple is the moment of
either force about any point in the line of the other.
Thus in Fig. (/') the forces P, P constitute a couple. The moment
is Pxp and is positive because the tendency is to rotate clockwise.
Theorem. Pn any system of fixes the algebraical sum of
the moments of the components about any point is equal to the
moment of the resultant about the point.
This important theorem is proved in books on mechanics. The re-
sultant moment may be called the moment of the system.
For example, two downward vertical forces, 8 ft. apart, each of 10
lbs., have moments of 60 and 20 ft. -lbs. about an intermediate point
K, distant 6 ft. and 2 ft. from their lines. The algebraical sum is
60 - 20 = 40 ft. -lbs. Now the resultant is 20 lbs. acting 2 ft. from K,
and its moment is 20 x 2 = 40 ft. -lbs., the same as before.
XXI
GRAPHIC STATICS
52i
5ccvt.
Examples on Problems 409 to 413.
*1. Copy the jointed frame double size with the supports placed at
the same level. The loads at the two upper joints, and the
supporting forces P and Q, are all vertical. Draw the force
diagram for the frame to a scale of J" to 100 lbs., and measure
/'and Q. Ans. 235 lbs., 405 lbs.
*2. Copy the braced cantilever double size. Then draw the force
diagram to a scale of j" to i cwt. , and measure the forces in the
four bars of the frame, distinguishing pulls from thrusts.
Ans. Pulls : 8.37, 20.1 ; thrusts: 6.68, 14. 2 cwt.
3. Draw a semicircle of 2J" radius with the four equal chords.
These chords and the diameter form a link polygon. Five
forces perpendicular to the diameter act at the joints and main-
tain equilibrium, causing a thrust of 10 lbs. in the long link.
Draw the force diagram, and determine the forces at the joints
and the tensions in the four short links.
4. In Figs. (<r) and (/>) above suppose the linear scale to be J" to
10", and the force scale " to 100 lbs., measure the distances
/CI/ and p. and the forces P, P, and calculate the moment of
P about K in (/7), and the moment of the couple in (/>).
Ans. 4S00 lb. -ft; 3880 lb. -ft.
522 GRAPHICS chap.
414. Problem. To find the moment about K of the
given force P of 7 lbs., the linear scale for the figure being
I" to 10".
Adopting Bow's notation, let the force be known as AB.
Selecting a suitable force scale (" to i lb. in the figure)
draw ab 7 units long to represent the given force AB.
Select any point as pole, at a distance os from ab re-
presenting unit force, or a force expressed by a simple
number. In the figure os measures 5 lbs. on the force
scale.
Join oa, ob. Through any point p on AB draw the two
lines, or links, parallel to oa, ob and of indefinite length.
Those links may be known as links A and B.
Through A' draw the line parallel to ab, to intersect the
two links A and B in A , B .
Then A B x os gives the moment of AB about K, these
lines being measured by the two scales, one as a length, the
other as a force.
^0^0 ao
Proof. Since the triangles pAJ3 M oab are similar, = ; or
AqBq x os = ab x pm = P x KM= moment of P about K.
Moment scale. A moment scale may be found on which
A Q B can be directly measured. Thus if A Q B were \", it
would measure 10" on the linear scale, and would represent
a moment of 10" x os lbs., that is 10" x 5 lbs. or 50 inch-
lbs. The moment scale is therefore \" to 10" x os lbs.,
that is \" to 50 inch-lbs. ; or more conveniently, V to 100
inch-lbs.
Measuring A Q B on this scale, the required moment is
found to be no inch-lbs.
Note 1. If K were any other point, the new intercept A B t) ,
measured on the moment scale, would give the moment of P about the
new A". The figure is thus a diagram of moments.
Note 2. The diagram is a link polygon for the force /"with respect
to the pole o. See Definition, Prob. 412.
If the links were secured each at a point, and hinged together at p,
they would form two bars of a frame, for which oab would be the force
diagram. The forces in the bars due to the load P at the joint p
would be found by measuring oa, ob on the force scale.
XXI
GRAPHIC STATICS
523
Linear /Scale
Force
Moment
i.e
or
414
< to 10"
1 4" to /lb.
M-" to io"*oslbs.
1 A" to I0"x5 2bs.
%. " to wo i?ich-lbs.
415. Problem. To find the moment about K of the
given couple 8 lbs., 8 lbs., the linear scale being given.
The linear and force scales of Prob. 414 are used.
Draw the two sides de, ef of the force polygon, to repre-
sent the given forces Q, Q, or DE, EF after Bow.
Take the pole o so that os is V, representing 4 lbs., thus
giving a moment scale of T V" to 10 inch-lbs.
Join oe, od, of Draw any link E parallel to oe, and
where this link meets the lines DE, EF, draw the two links
D, F of indefinite length, parallel to od, of.
Draw through K the line parallel to de to cut the two
links D and F in D , F .
Then D Q F gives the moment of the couple about K.
Measured on the scale of T V' to 10 inch-lbs., the moment
is seen to be 117 inch-lbs.
Note 1. As before, the link polygon is a diagram of moments.
The links D and /'are parallel, and the intercept -D ( f is of constant
length for all positions of A', illustrating the well-known theorem that
a couple exerts the same turning moment about any point.
Note 2. Owing to the limited space the figures in the book are
necessarily small. But the student, with a sheet of drawing-paper at
his disposal, should, where possible, select scales which give diagrams of
ample size.
524 GRAPHICS chap.
416. Problem. - - To find the resultant of a given non-
concurrent system of forces in one plane.
Let P, Q, M, IV ox AB, BC, CD, >E be the forces.
First, to find the magnitude, direction, and sense of the re-
sultant. Draw the four circuital sides ah, be, cd, de of the
force polygon to correspond with the given forces. Join ae
and direct it non-circuitally. Then the vector ae, when
localised, will represent the resultant.
Next, to find a point in the line of the resultant. Choose
any pole o, and join oa, oh, oc, od, oe.
Start from any point / on AB. Through p draw the
link/;- or A of indefinite length 'parallel to oa; draw also
the link pq or B between AB, BC, and parallel to ob.
Draw the link qm or C, between BC, CD, and parallel to
oc, Draw the link inn or D, between CD, DE, and parallel
to od. And draw the closing link nr or E parallel to oe to
meet the first link A in r.
Then r is the required point. A line through r, vector-
ally equal to ae, represents the resultant completely.
Reason. If the link polygon were a hinged frame with the forces P,
Q, M, N, and R reversed in sense, acting at the joints, it will be found
on examination that oabedeo would be its force diagram, and that each
of the joints, and therefore the whole frame, would be in equilibrium.
Note I. Rule for drawing the links. For a balanced system of
forces, or for a system to which the resultant has been added, observe
that in Bow's modified notation there are the same number of letters as
forces, and that each letter is associated with two, and only two, forces.
And in any complete or closed link polygon the links A, B, . . . are
parallel to oa, ob, . . . and have their ends on the pairs of forces
which have A, B . . . common.
For example, the link B is parallel to oh and terminated by AB, BC
which have B common. Attention to this rule will prevent mistakes.
417. Properties of the link polygon. Conditions of
equilibrium of coplanar forces.
(a) Partial resultant. The resultant of any portion of the system
for which the vectors are consecutive in the force polygon can readily
be found.
Thus the resultant of BC, CD is represented by a vector equal to
bd, localised through the point t where the links B, D intersect.
Observe that Z?,Z> are the first and last letters in the sequence B C, CD.
XXI
CRAI'IIIC STATICS
525
416
N /E
(/>) Moment of the system about any point A". Througli K draw
a line parallel to ae to meet the links A, E in A w E w Draw os per-
pendicular to ae.
Then the required moment = A Q E x os, the scales being determined
as in Prob. 414.
(c) Partial moment. The moment of any part of the system for
which the vectors are added in sequence as in (a) is easily found.
Thus to obtain the moment of BC, CD about A", draw a line througli
A" parallel to bd, and draw os' perpendicular to bd.
Then the required moment = B Q D x os'.
(d) Parallel forces. If the forces are all parallel the vector polygon
becomes a line ; the lengths of os, os become equal ; and the lines
through A", A"' are parallel to the system.
Thus for parallel forces the link polygon is a very useful diagram of
moments, the scale being the same for all the intercepts.
(e) Conditions of equilibrium of a system of forces in one plane.
The necessary and sufficient conditions are two, viz.
1. The force polygon must close.
2. The link polygon must close.
The first ensures that there shall be no resultant force, though there
might be a couple.
The second ensures that there shall be no couple, since when this
condition holds, the moment of the system about any point is zero.
By the closing of the link polygon is meant that the last link shall
intersect the first link on the first force.
We shall now give some simple problems illustrating the application
of the above general principles to special cases.
526 GRAPHICS chap.
418. Problem. Three parallel forces L, M, N of given
magnitudes act as shown ; (a) find the parallel forces X
and Y along the given lines which will balance them ; (b)
find the resultant of L, M, N ; (c) find the moment of
the given forces about K ; (d) find the moment of Y and
L about K.
The scales are given in the figure, the moment scale being derived
as explained in Prob. 414.
Employing Bow's modified notation, the letters AB, BC, CD, DE,
EA, forming a cycle, are appended to the five balanced forces
LMNXY.
{a) To find X and Y. Draw the three sides ab, be, cd of the force
polygon to correspond with the known forces AB, BC, CD.
Select a pole 0, and join oa, ob, oc, od.
Draw the links yl, Im, mn, nx (in Bow's notation named A, B,
C, D) parallel to oa, ob, oc, od.
Draw xy {i.e. the dosing link E). And draw oe parallel to xy, thus
determining the closing sides de, ea of the force polygon.
The forces X and Y, that is DE, EA, are now found by measuring
de, ea on the force scale. They are 8 lbs. and 38 lbs., in the senses
given by the sequences de, ea, that is they both act upwards.
[b) To find the resultant of AB, BC, CD. The first and last letters
of this sequence are A and D.
The resultant acts through r where the links A and D intersect.
And it is found by measuring ad to be a dozunward force of 46 lbs.
(r) To find the moment of A B, BC, CD about K. Again observe
that the first and last letters of this sequence are A and D.
Draw through K a line parallel to ad, to meet the links A and D
in A , D .
Then A D represents the required moment, which measured on the
moment scale gives the value 61 ft. -lbs.
Note 1. os was taken 50 lbs. on the force scale, which in conjunction
with the linear scale leads to the easy moment scale of 1" to 100 ft.-
lhs. See Prob. 414.
(d) To find the moment of EA, AB about K. The first and last
letters are E and B.
Draw through K a line parallel to eb, to cut the links E and B in
E B .
Measuring E^B^ on the moment scale, the required answer is 47 ft.-
lbs.
Note 2. The student of applied mechanics will recognise in Fig.
418 a diagram of bending moments for a beam supported at the ends
and loaded at intermediate points. See also Fig. 420.
XXI
GRAPHIC STATICS
527
L=31 M=60 N=45
A \B K b\c c\d
Linear scale h'bol' Force scale 3 /i6 'to to lbs.
Moment scale '/z "to J \ 50 lbs. ? i.e. 1 "to WO ft. lbs
L g M N
h
A\D
AB
419
419. Problem. HKLM is a square of 2h" side. Two
forces of 9 units each act from H to K, and from L to M,
forming a couple. A third force of 4 units acts from L to
K. Find their resultant.
The scales for the figure are j" to i" and " to I unit of force.
The given forces are lettered AB, BC, CD. The required resultant
will be AD.
Draw the three circuital sides a&, be, cd of the force polygon, and
the non-circuital closing side ad.
The pole o is chosen at the intersection of ab and cd.
The link polygon is begun at A' on AB. The first link A is a line
through K parallel to oa. The second link B is the point K. The
third link C is the line AW parallel to oc. The closing link D is a
line drawn through N parallel to od to meet the first link, which it does
in A 7 "; so the link D reduces to the point A T .
Thus the required resultant AD passes through A\ a point on LM
produced, distant by measurement 3.12" from M, and is a downward
force of 4 units as given by ad.
52 8 GRAPHICS chap.
420. A uniform horizontal rod HK 6 ft. long and weigh-
ing l. 1 , lbs. is hinged at H, and a downward vertical force
of 3i"lbs. is applied 2 feet from the hinge. Where must
an upward vertical force of 2 lbs. be applied to maintain
equilibrium ; and what will be the pressure on the hinge ?
The scales for the figure are \" to i' and \" to I lb.
The weight of the rod may be supposed to act at its middle point.
Let AB, BC, CD, DA denote the four forces as in the figure, of
which the first two are given completely, and we require to find the
line of the third and the magnitude of the fourth.
Draw the three circuital sides ab, be, cd of the force polygon, to re-
present the three known magnitudes. The non-circuital closing side
ad gives a downward force of 3 lbs. for the pressure on the hinge, this
being the resultant of the other three forces.
Take any pole o and join oa, ob, oc, od.
Draw the links A, B parallel to oa, ob, and the two closing links C,
D parallel to oc, od to meet at t.
The vertical line tL through t is the required line of action of the
supporting force, and HL measures 5.75 feet.
Examples on Problems 414 to 420.
*1. Copy the diagram double size, and let the linear scale then be
1" to the foot. Take for the force scale J" to 10 lbs., and let
the polar distance os be 2", representing 40 lbs.
(a) Find the resultant of M, P, Q, N. Aus. 18 lbs. upwards
6.6 feet from Y.
(b) Find the moment of Af about A". Ans. - 54 lb.-ft.
(<) Find the moment of the couple P, Q about A'.
Ans. 80.6 lb.-ft.
{d) Find the moment of M, P, Q, N about K.
Ans. -82. 8 lb.-ft.
((.') Find the forces X, Y which balance M, P, Q, A\
Ans. 36.9 lbs. down ; 18.9 lbs. up.
*2. The jib HJ of a 3-ton crane is inclined at 57 to the horizontal,
and the tie rod ///at an angle of 27 . Find the thrust in the
jib and the pull in the tie. Ans. 5.35 tons, 3.27 tons.
If a back stay IK be added inclined at 45 , and attached to
the end of a horizontal strut ^/A', find the counter-balance weight
IV required at K to balance the load on the crane about J.
Find also the tension in the back stay and the thrust in the
crane post //. Ans. 2.91 tons; 4.12 tons; 1.43 tons.
Note. The counter balance weight IV should be found by
two methods ; first by drawing the force diagram for the frame ;
next by drawing a link polygon for the three external forces W,
the load, and the vertical supporting force aty.
XXI
GRAPHIC STATICS
529
35 1-5 ii?S
H A\B\C
?
a
5.
On a horizontal line OX mark off towards X the lengths OA =
0.5", OB =1.1", 0C= 2.0", 0Z> = 2.5", and above the line set
off the^ angles OAP= 3S , OBQ=-jo, OCM=no, and ODN
= 120 . Suppose forces of 320, 145, 570, and 416 lbs. to act
respectively along PA, QB, MC, and ND. Find the resultant
force, and the resultant moment about 0, if the linear scale is
|. Measure and write down (a) the magnitude of the resultant ;
(/') the angle which its line of action makes with OX ; (c) the
distance from O (to scale) of the point where the resultant cuts
OX; (d) the resultant moment of the system about O.
Ans. (a) 1220 lbs., (b) 94. 3 , (c) 7.T,", (d) 8S60 inch-lbs.
Draw a rectangle ABCD, making AB= 3 .6", BC=\.^'.
Forces each of 6.5 lbs. act along AB and CD thus constituting
a couple. Determine graphically and measure the moment
of this couple. Ans. 9. 1 inch-lbs.
In Ex. 4 determine graphically the forces of a couple which,
acting along the short sides of the rectangle, shall balance the
given couple. Ans. 2.53 lbs. in the senses AD and CB.
In Ex. 4 let two additional forces each of 3.7 lbs. act along CB
and DA. Determine graphically the resultant of the four
forces, and find the moment of the system about a point A"
inside the rectangle, distant 0.9" from both AB and CD.
2 M
530 GRAPHICS chap.
421. Centre of gravity. Two examples are now given of
the determination of the centre of area of a plane figure,
often termed its centre of gravity.
The first case represents the cross section of a cast-iron
beam, symmetrical about a vertical axis YY. The figure
is divided into three rectangles, the areas of which are
calculated from the data, supposed given.
Now through the centres of these rectangles draw parallel
vectors AB, BC, CD in any direction other than that of
YY (but at right angles to FKfor convenience), and let the
magnitudes of the vectors be proportional to the calculated
areas. Draw the vector polygon abed, and a link polygon
to any pole o, and thus determine the resultant vector AD,
passing through the intersection of the closing links.
The centre of area required is the point G, where the
resultant vector AD intersects YY, the axis of symmetry.
In the second example, that of a section of angle iron,
there being no axis of symmetry, it is necessary to draw two
link polygons, the parallel vectors, AB, BC, of one having
a direction differing from those, A'B', B'C, of the other.
The intersection of the resultants AC, A'C gives G, the
required centre of area of the section.
Examples. 1. In the upper figure opposite let the dimensions of
the top flange be 4" by 1^"; of the bottom flange 9" by i-J"; and
of the vertical web 10" by 1". Determine the distance of the
centre of area from the top. Am. 8".
2. In the lower figure let the angle iron be 4" by z\" outside, and
the thickness i". Find the centre of area. Aiis. 1.42" and
0.67" from the top and left sides.
3- A uniform straight wire 4^" long is bent at right angles at a point
1 1" from one end ; find its centre of gravity. Ans. j" and 1
from the long and short sides.
4. A straight wire 6" long is bent at points distant 1.2" and 3.2
from end, forming three straight lengths, the first and last being
parallel, and both making 6o Q with the middle segments. Find
the centre of gravity of the wire.
5. A uniform wire 6" long is bent into the form of a semicircle with
the diameter. Find its centre of gravity.
6. Find the centre of area of the segment of a circle of 2" radius,
the chord being 3^" long.
XXI
GRAPHIC STATU S
53i
T)
Y
1
!-H - - -
C
1
1
1
1
G
B
k
A \
532 GRAPHICS chap.
422. Problem. Known forces are balanced by two
others, the line of action of one of which, and a point in
the line of action of the other, are given. To find the
magnitude and sense of the first balancing force, and the
magnitude, line of action, and sense of the other.
Let AB, BC, CD be the known forces ; and let YYbe
the given line of action of one of the balancing forces, which
call DE, and P the given point in the line of action of the
other ; this latter, on Bow's system, must be labelled EA.
To find the magnitude and sense of DE, and the magni-
tude, line of action, and sense of EA.
Draw the force polygon so far as the data will admit,
viz. the sides ab, be, cd, and an indefinite side through d
parallel to DE. The solution consists in finding the point
e in this line.
Take any pole o, and join oa, ob, oc, od. Draw the link
polygon with respect to this pole, beginning at the given
point P. Although the line of action of the force at P is
unknown, yet P is a point on it, and we may therefore
begin the link polygon at P. Call y the point where the
fourth link D meets YY. Join Py in order to close the link
polygon. Draw oe parallel to Py, and join ea to close the
force polygon. Then de gives the required magnitude and
sense of the balancing force DE along YY; and ea gives
the magnitude, line of action, and sense of EA, the balan-
cing force which acts through P.
This problem occurs in connection with roof trusses
provided with expansion rollers at one end, and subjected
to the pressure of the wind.
423. Problem. Any number of known forces are
balanced by three others which act along given lines. To
find the magnitudes and senses of the three balancing
forces or reactions.
Let AB, BC, CD be the given forces, and XX, YY,
ZZ the given lines along which the balancing forces act. To
find the magnitudes and senses of the latter.
XXI
CRAI'HIC STATICS
533
Denote the required forces by DE, EF, FA as shown.
Draw the force polygon so far as the data allows ; that is
draw ab, be, ed, and the indefinite sides through d and a
parallel to DE and AF The problem is solved when we
have found e and /on these two lines.
Take any pole o, and join oa, ob, oe, od. Begin the link
polygon at one of the intersections of the lines X, Y, Z, say
at p. The link F then reduces to a point. Draw the links
A, B, C, D, terminating at x on XX.
Close the link polygon by drawing px which is the link E.
Then close the force polygon by drawing oe parallel to px or
E, and ^/"parallel to EF.
Then de, ef, fa taken circuitally give the required magni-
tudes and senses of the balancing forces DE, EF, FA
acting along XX, YY, ZZ.
534
GRAPHICS chap.
424. Miscellaneous Examples.
*1. The figure shows a buttress subject to the forces of its own
weight and a thrust of 2800 lbs. at its upper end. Draw the
" line of resistance " for the structure.
Hint, This is a link polygon for the forces, beginning at
the point of intersection of the forces 2800 lbs. and 400 lbs.
Compare Prob. 412.
*2. A heavy uniform wire is bent into the form ABCD, and is sus-
pended by a string attached to the point A. Draw the direction
of the string. (1892)
*3. The directions and magnitudes in lbs. of five unequal forces
acting at a point a are given. Determine the direction and
magnitude of their resultant. 0884)
*4. Five given forces act as shown at a point. Determine by
construction two forces acting along the given dotted lines which
will keep the point in equilibrium. Write down the magni-
tudes and indicate the directions of these forces. Use a scale
ofi"=ioolbs. (1887)
*5. Resolve each of the given forces P and Q (P~- 7 lbs., = 9 lbs.)
along and perpendicular to the given line AB, and write down
the resultant force in each direction. (1S85)
*6. A uniform beam AB, weight 44 lbs., is suspended by two equal
strings from its extremities to the point 0. A weight /' ( = 1 3
lbs.) is hung on the beam in the given position. Determine the
position of equilibrium of the system.; ('891)
*7. A weight of 5 tons is suspended from a, the apex of a triangle
formed of two bars ba, ca fixed in a vertical wall. Determine
and write down the stresses in the bars ba, ca. (1893)
8. The wire passing round the top of a telegraph pole is horizontal,
and the two directions make an angle of 1 io with one another.
The pole is supported by a wire stay inclined at 45 to the hori-
zon. Given the tension of the telegraph wire to be 200 lbs.,
find that of the stay. Ans. 324 lbs. (1896)
9. ABCD is a square, the angular points being lettered in order.
Two forces, of 10 units each, act from .-/ to B and from C to
D, forming a couple. A third force of 15 units acts from C to
A. Find their resultant. (1892)
10. Draw six lines oa, ob, oc, od, oe, of, radiating from a point o,
any two adjacent lines including an angle of 6o Q . These six
lines are respectively the lines of action of forces of 80, 100, 90,
60, 120, and 50 lbs. all acting towards except those along oa
and of, which act away from 0. Determine and write down the
magnitude and direction of the resultant of the forces.
Ans. A pull of 1 15 lbs., making 4.3 and 55.7 with oc and od.
XXI
GRAPHIC STATICS
535
Cvfw //ie /imwvs dwMe size
536 GRAPHICS
CHAT.
*11. Determine the line of action, and write down the magnitude
of the resultant of the five given parallel forces acting in one
plane in the directions shown by the arrows. (1888)
*12. Four vertical forces act downwards as follows :
At A 10 lbs., at B 18 lbs., at C 16 lbs., and at D 12 lbs.
Determine the position of their resultant.
Supposing the two forces at B and C only to act downwards,
determine the values of two vertical forces acting upwards
through A and D so as to make equilibrium with those through
B and C. Employ for the force polygon the scale o. 1" to 2
lbs. (1895)
*13. Find (and write down) the moment in foot-tons of the result-
ant of the pairs of parallel forces, A and B, C and D, with
regard to the point E. Scale of forces, 0.25" per ton ; scale of
distances, o. 1" per foot. ( J S94)
*14. A force of 14! lbs. acts along the given line ab. Determine
by construction what force acting along cd will have the same
moment about P. (1886)
*15. A uniform rod AB, weighing 53 lbs., is pivoted at A. If a
force P of 32 lbs. is applied at C, where must a parallel force of
41 lbs. be applied to maintain equilibrium?
*16. Obtain by construction a line representing the moment of the
resultant of the two given forces P, Q, about the point 0, using
a scale of |"= 1 lb., and linear scale full size. (1888)
*17. / > 1 =i5o lbs., P., = 200 lbs., P 3 =i20 lbs., are three forces
acting in the direction indicated by the arrow-heads. Find by
the funicular polygon the resultant moment of the three forces
acting round the point S.
Scale of forces, 1 00 lbs. = I inch. Scale of lengths, 50 feet
= 1 inch. (1897)
18. A bar of uniform thickness inclined at an angle of 30 to the
horizontal, with one end against a wall, rests across a rail at a
point 2 feet away from that end. Find the length of the bar if
the rail and wall are both smooth. Ans. 5 feet 4 inches.
Hint. The three forces which act on the bar are ( 1 ) the
force from the wall, which is in a direction at right angles to
the wall, (2) the force from the rail which is at right angles to
the bar, (3) the weight of the bar which acts vertically through
its middle point. Now use Theorem "] , Art. 401.
19. Three forces of 11, 19A, and 26 lbs. act at a point Pin such
directions that their resultant is nil. Draw lines representing
the forces in direction and magnitude. (1886)
XXI
GRAPHIC STATICS
557
Cbftv /he fhures dot idle- sue-
A
B
^
^
% "s
^ s
r
'
'
1
I
11
A
Tons 4
B
C D,
Tons 2 Tons 5lTons
13
,E
P
P
15
C
C
D
B
S
17
?*
APPENDIX I
SCIENCE AND ART DEPARTMENT EXAMINATION
May 1S99
Advanced Stage
Instructions
Only eight questions are to be attempted.
Plane Geometry.
*21. C is one vertex of a triangle ; (J is the centre of the circum-
scribed circle ; the centre of the inscribed circle. Draw the triangle.
(22)
22. Draw an indefinite line PC. At any point in it, N, draw a
line perpendicular to BC, and set off from N on the perpendicular,
above and below BC, lengths NP, NP 2 , each equal to 2.4 inches.
BC is the axis of a parabola, and P, P 2 are points on the curve. Calling
A the vertex of the parabola, draw the curve such that the area
bounded by the double ordinate PP 2 and the portion of the curve
/'A P., shall be 4 square inches in area. (20)
*23. Two equal elliptic wheels, A and B, are in contact at 0.
Their foci are : those of A, I\, and E, ; those of B, f v and /.,. The
wheel A is driven by the wheel B ; and the wheels rotate round fixed
axes at the foci I\ andy^ respectively. A pin is attached to A at its
focus F 2 . Draw a diagram representing the vertical heights of the pin
above the line CD during a complete revolution of the wheels, the pin
at the commencement of the revolution being at the point F 2 on the
diagram. Take for abscissae \ inch to represent T Vth of a complete
revolution ; and for ordinates the actual heights of the pin corre-
sponding, above CD. The position should be shown for, at least,
every T Vth of a complete revolution. (22)
Solid Geometry.
*24. a!>, db' are the projections of a line AB ; I'h, hgzxe. the traces
of a plane. Draw the projections of a line in the plane, meeting AB,
and making with it an angle of 35 . (24)
AI'PEN. I
EXAMINATION PAPER
539
The diagrams (ejccenbN?26)tobeprikedoff or (7899)
accurate/ f transferred to the paper.
Q
G
c
O
A
/ \
/ \
C / \ 7)
540 ADVANCED STAGE appen.
25. A cube, inscribed in a sphere of i^ 5 inches radius, has two
adjacent faces inclined at 30 and 70 respectively, to the horizontal
plane. Draw the solids in plan and elevation. (24)
*26. Draw an equilateral triangle abc (see diagram, which is not
drawn to scale) of 4.2 inches side, and set off lengths aA, aB, cF, etc. ,
1 inch distant from each vertex, along the sides. The figure ABCDEF,
so formed, is a horizontal section of a regular octahedron, lying with
one face on the horizontal plane. Draw the octahedron, showing on it
the outline of the section in plan, and its height above the horizontal
plane in elevation. (26)
*27. A and B are the scales of slope of two planes. Draw the plan
of a sphere of I inch radius, resting on the horizontal plane, and touch-
ing the two planes, both of which pass over the sphere. Show the points
of contact with the planes, and write down their figured heights.
Unit = o. 1 inch. (24)
*28. ad ', bb', cc' are the projections of three points A, B, and C.
Find the projections of a fourth point D, distant 2j inches from each
of the points A, B, and C. Complete the tetrahedron formed by
joining the four points. (22)
29. The vertex of a cone is 1.5 inches above the horizontal plane,
and its axis is inclined at 45". Its generating lines make an angle
of 30 with the axis. Determine the scale of slope of a plane tangent
to the cone, and inclined at 6o u . Show the line of contact on the cone
in plan. Unit = 0.1 inch. (24)
*30. The diagram represents a cone AB V, lying on a block DEFG
whose thickness is DD 2 . Draw on the plan the outline of shadow
thrown by the solids, one on the other, and on the horizontal plane
of projection. Show also on the plan the limit of light and shade on
the cone. The arrows indicate the direction of the parallel rays of
light, inclined at 45 to the xy line in plan and elevation. (30)
*31. Draw the solids of Question 30 in isometric projection. The
vertical isometric planes to be taken parallel to the planes DE and
DG, and the vertical through D nearest to the observer. An isometric
scale must be employed. (24)
Graphic Statics.
Alternative and Optional.
32. A right truncated prism has for base an equilateral triangle of 2
inches side. The three edges perpendicular to the base are respec-
tively 2 inches, 1.75 inches, and 1.25 inches in length. Find, by
graphic construction, a line representing the cubic contents of the
solid, to a unit of 1 inch. (20)
*33. ABCD is a horizontal rigid bar, hinged at A, loaded ati? with
40 lbs., at 6' with 50 lbs., and retained in place by a cord DE (passing
over a pulley) attached to a pin at D. Find the stress in the cord DE.
Employ the funicular polygon, using, for the scale of loads, J inch to
represent 10 lbs. ( 2 4)
EXAMINATION PAPER
541
(1899)
Note. The figures are reproduced half size.
APPENDIX II
DEFINITIONS AND THEOREMS OF PURE SOLID GEOMETRY
Definitions
Definition 1. & plane is a surface such that any two points being
taken in it, the straight line joining them lies wholly in the surface.
Definition 2. Parallel planes are such as do not meet each other,
though produced.
Definition3. A straight line is parallel to a plane when the two
do not meet each other, though produced.
Definition 4. A straight line is perpendicular to a plane, when it
is perpendicular to every straight line which meets it in that plane.
Definition 5. Two planes are perpendicular to each other when
any straight line drawn in one, perpendicular to the intersection of the
planes, is perpendicular to the other plane.
Definition 6. The orthographic projection of a point on a plane
is the foot of the perpendicular from the point to the plane. The per-
pendicular is called a projector, and the plane is called thep/ane of pro-
jection.
Definition 7. The orthographic projection of a given line on a
plane is the line generated by the foot of a perpendicular to the plane,
which perpendicular moves so as always to intersect the given line.
The surface generated by the moving perpendicular is called the
projecting surface. When the line which is projected is straight, the
projecting surface is called the projecting plane.
Definition 8. The inclination of a straight line to a plane is the
angle between the line and its orthographic projection on the plane.
Definition 9. The angle between two planes, called a dihedral
angle, is measured by the angle between two straight lines, drawn from
a point in their intersection, each perpendicular to the intersection, and
lying one in each plane.
appex. ii DEFINITIONS 543
Definition 10. The inclination to each other of two straight lines
in space which do not intersect is measured by the angle between two
lines from any point, respectively parallel to those lines.
Definition 11. A solid ox polyhedral angle is formed when three or
more planes meet in a point. It consists of as many plane angles, and
also of as many dihedral angles as there are planes.
Definition 12. A solid is that which has length, breadth, and
thickness. It is completely bounded by a surface, or by surfaces, which
may be plane or carved.
Definition 13. A. polyhedral is a solid bounded by plane surfaces,
called the faces, which meet in straight lines called the edges.
Definition 14. A prism is a polyhedron of which the side faces
are parallelograms, and the two end faces, or bases, are similar and equal
polygons in parallel planes. The line joining the centres of the bases
is called the axis of the prism.
If the axis be perpendicular to the base, the prism is said to be a
right prism ; if not, it is said to be oblique.
The perpendicular distance between the bases is called the altitude.
Definition 15. A pyramid is a polyhedron, one" face of which,
called the base, is a polygon, the other faces being triangles which have
a common vertex.
The common vertex of the triangles is called the vertex of the
pyramid, and the line joining the vertex to the centre of the base is
called the axis of the pyramid. If the axis be perpendicular to the
base, the pyramid is said to be a right pyramid ; if not, it is said to be
oblique. The perpendicular distance from the vertex to the base is
called the altitude.
Definition 16. A pyramid having a triangular base is called a
tetrahedron.
Definition 17. A polyhedron is said to be regular when its faces
are similar, equal, and regular polygons, and all its dihedral angles are
equal to one another.
Definition 18. A regular tetrahedron is a solid having four equal
and equilateral triangles for its faces.
Definition 19. A cube is a solid having six equal squares for its
faces.
Definition 20. A regular octahedron is a solid having eight equal
and equilateral triangles for its faces.
Definition 21. A regular dodecahedron is a solid having twelve
equal and regular pentagons for its faces.
Definition 22. A regular icosahedron is a solid having twenty
544 PURE SOLID GEOMETRY appen.
equal and equilateral triangles for its faces, and all its dihedral angles
equal.
Definition 23. A surface of revolution is the surface generated
by the rotation of a line (straight or curved), about a fixed straight line,
to which it is supposed to be rigidly connected.
The fixed straight line is called the axis, and the rotating line the
generator.
Definition 24. A conical surface is that generated by a straight
line which moves so as to always pass through a fixed point, and to
intersect a fixed curve in space.
The fixed point is called the vertex. The fixed curve is called the
directing curve.
Definition 25. A conical surface of revolution is the surface
generated by one of two intersecting straight lines, rotating about the
other as axis, the lines being supposed rigidly connected together.
The point of intersection of the lines is called the vertex, and the
complete surface consists of two portions extending indefinitely, one on
each side of the vertex.
Definition 26. A right circular cone is the solid generated by
the revolution of a right-angled triangle about one of the sides con-
taining the right angle as axis. The circle generated by the other of
the sides containing the right angle is called the base of the cone.
Definition 27. A cylindrical surface is that generated by a straight
line which moves so as to be always parallel to a fixed straight line,
and to intersect a fixed curve.
Definition 28. A right circular cylinder is the solid generated by
the revolution of a rectangle about one side as axis.
The two sides of the rectangle which are perpendicular to the axis
generate circles, each of which is called a base of the cylinder.
Definition 29. A sphere is the solid generated by the revolution
of a semicircle about its diameter as axis. The centre of the sphere is
the point which is equidistant from all points in the surface.
Theorems
Theorem 1. The plane which contains two parallel straight lines
will also contain any third straight line which intersects them.
Theorem 2. A plane can be found which shall contain two inter-
secting straight lines.
Cor. A plane is determined by three intersecting straight lines AB,
BC, CA not concurrent, or by three points A, B, C not collinear.
Theorem 3. If two planes cut each other, their intersection is a
straight line.
II THEOREMS 545
Theorem 4. If a straight line be perpendicular to each of two
other straight lines at their point of ' intersection t it will be perpendicular
to the plane containing the t-wo lines.
Theorem 5. If one of two parallel straight lines be perpendicular
to a plane, the other line will also be perpendicular to the plane.
Theorem 6. If a straight line be perpendicular to a plane, every
plane containing the line -will be perpendicular to that plane.
Theorem 7. The orthographic projection of a finite straight line
on a plane is the straight line joining the projections of its ends.
Theorem 8. If three or more straight lines which meet at a point
are each perpendicular to the same straight line, they are in one plane.
Theorem 9. If two or more straight lines are perpendicular to the
same plane, they are parallel to one another.
Theorem 10. If two or more straight lines are parallel to the same
straight line, they are parallel to one another.
Theorem 11. If two straight lines -which meet are respectively
parallel to two others which meet, but are not in the same plane, the
first two and the other two will contain equal angles. Also the plane
containing the first two is parallel to the plane containing the other two.
Theorem 12. Planes which are perpendicular to the same straight
line are parallel to one another.
Theorem 13. If two or more pa7-allel planes be cut by another
plane, the lines of intersection are parallel to one another.
Theorem 14. If two or more straight lines be cut by parallel
planes, they will be cut in the same ratio.
Theorem 15. If two intersecting planes be each perpendicular to
a third plane, their intersection will also be perpendicular to that plane.
Theorem 16. If two planes which meet be cut by a third plane
-which is perpendicular to their line of intersection, the lines in which
the third plane cuts the other two will both be perpendicular to the inter-
section of the other two.
Theorem 17. If two or more straight lines be parallel to one
another, their projections on any plane will also be parallel to one another.
Theorem 18. If a finite straight line be parallel to a plane, the
lengths of the line and its projection on the plane are equal to each other.
If the line be inclined to the plane, the length of the projection is less
than the length of the line. If the line be perpendicular to the plane,
its projection is a poi>it.
Theorem 19. If a straight line be divided into two or more seg-
ments, their projections on any plane (iwt perpendicular to the line) are
in the same ratio as the segments themselves.
2 N
546 PURE SOLID GEOMETRY ArPEN.
Theorem 20. If two planes intersect, and a straight line perpen-
dicular to one be projected on the other, the projection is perpendicular
to the intersection.
Theorem 21. If two straight lines be perpendicular to each other,
their projections on any plane parallel to one of them will also be perpen-
dicular to each other.
Theorem 22. If the projections of two straight lines on any plane
be perpendicular to each other, and one of the lines is parallel to the plane
of projection, the two lines are perpendicular to each other.
The Sphere and the circular Cone and Cylinder.
Theorem 23. The projection of a sphere on any plane is a circle
of diameter equal to the diameter of the sphere, the centre of the circle
being the projection of the centre of the sphere.
Theorem 24. Any plane section of a sphere is a circle, the centre
C of which is the foot of the perpendicular from the centre of the sphere.
Note. If the plane contain O, the section is called a great circle ;
other sections are small circles.
Theorem 25. The intersection of two spheres is a circle whose
plane is perpendicular to the line joining the centres of the spheres, the
centre of the circle being in this line.
Theorem 27- The tangent plane to a sphere, centre O, at any
point P on its surface is perpendicular to OP.
Theorem 28. A line tangential to a sphere, centre 0, at any point
P on its surface is perpendicular to OP.
Theorem 29. If two spheres touch, the point of contact is in the
line joining their centres.
Theorem 30- The section of a cone or cylinder by a plane perpen-
dicular to the axis is a circle whose centre is in the axis.
Theorem 31. The intersection of a cone or cylinder by a sphere
whose centre is in the axis consists of two circles with their planes per-
pendicular to the axis and their centres in the axis.
Theorem 32. /// a cone or cylinder a sphere can be inscribed
which lias any given point on the axis as centre, or which passes
through any given point P on the surface. The curve of contact is a
circle whose centre is in the axis and whose plane is perpendicular thereto.
Theorem 33. A cone of indefinite length may be defined by any
two inscribed spheres, or by one such sphere and the vertex.
Theorem 34. A cylinder of indefinite length may be defined by
any two inscribed spheres, or by one such sphere and the axis.
n THEOREMS 547
Theorem 35. The projection of a cone or cylinder of indefinite
length on any plane consists of two lines which touch the projections of
any two inscribed spheres.
Note. For the cylinder, if the plane be perpendicular to the axis,
the projection is a circle which is an edge view of the surface. For the
cone, if the projection of one inscribed sphere fall within that of the
other, the surface has no definite form of projection. For illustrations
of Theorem 35, see the figure on page 321.
Theorem 36. The projection of a cone or cylinder {or of any other
surface of revolution) maybe determined as the envelope of the projections
of its inscribed spheres.
Theorem 37- Any plane containing the axis of a cone or cylinder
cuts the surface in a pair of generating lines.
Theorem 38. The tangent plane to a cone or cylinder at any point
P on its surface is perpendicular to the plane which contains P and the
axis. The tangent plane has line contact with the surface, this line
being the generator through P. The axial plane through P cuts the sur-
face in the line of contact.
Theorem 39. Any tangent plane to a cone or cylinder touches all
inscribed spheres.
Theorem 40. Any plane which contains the vertex of a cone and
which touches an inscribed sphere also touches the cone.
Theorem 41. Any plane zvhich 'is parallel to the axis of a cylinder
and which touches an inscribed sphere also touches the cylinder.
Theorem 42. If two cones, or two cylinders, or a cone and cylinder
circumscribe the same sphere, the intersection of their surfaces is a pair
of ellipses whose planes are perpendicular to the axes of the surfaces.
This is found to be a very useful theorem.
Theorem 43. A common tangent plane to two cylinders can in
general be found {a) when their axes are parallel ; (b) when they cir-
cumscribe the same sphere.
Note. The tangent plane may be determined as the plane which is
parallel to the axes, and which in (a) touches any two spheres inscribed
one in each cylinder ; and in (b) touches the common sphere.
Theorem 44. A common tangent plane to two cones can in general
be found (a) when their axes are parallel and their vertical angles equal ;
(b) when they have a common vertex ; and (c) when they both circum-
scribe the same sphere.
Note. The tangent plane may be determined as the plane which in
(a) contains the two vertices and touches any inscribed sphere; in (b) con-
tains the common vertex and touches two spheres inscribed one in each
548 PURE SOLID GEOMETRY appen. 11
cone ; and which in (<) contains the vertices and touches the common
sphere.
Theorem 45. A common tangent plane to a cone and cylinder can
be found when they circumscribe the same sphere.
Note i. The tangent plane may be determined as the plane which
contains the vertex of the cone ; is parallel to the axis of the cyliftder,
and which touches the common sphere.
Note 2. In theorems 43, 44, and 45 there may be impossible cases.
Moreover the conditions stated are not exhaustive.
Theorem 46. The normal from any point to the surface of a cone
or cylinder lies in the plane determined by the point and the axis, and
is perpendicular to one of the generators in which this axial plane cuts
the surface.
Theorem 47. If a sphere touch a cone or cylinder, the point of
contact lies in the generator determined by the axial plane through the
centre of the sphere.
Theorem 48. Two cones, two cylinders, or a cone and cylinder
which touch may have line contact or point contact.
If they have line contact their axes must intersect or be parallel ; and
the line of contact is the generator determined by the common axial plane.
If they have point contact at P, the common tangent plane at P con-
tains the two generators through P.
Theorem 49. The shortest path on a sphere between two points is
the smaller arc of the great circle through them.
INDEX
The references are to pages
Abscissa, 150
Addition, graphic, 487
Adjustment of set-squares, 2
Alignment, true, 2
Amplitude, 128
Analytical geometry, 156
Angle between line and plane, 249
between traces of a plane, 246
between two lines, 246, 248
between two planes, 249, 284
chord of, 19
how to measure an, 22
pitch, 476, 478
to set off an, 22
Angles, tables of sines, etc. , 20
trihedral, 179, 466
ways of denning, 18
Appendix I., 538
II-. 542
Archimedian spiral, 124
Arc, length of a circular, 112
Area, centre of, 530
mean ordinate of, 44
of polygon, 36, 492
Arithmetic, graphic, 486
Asymptotes, 118, 126
of hyperbola, 104
Auxiliary circle, 78
circle, major, 78
circle, minor, 78
elevation, 196, 197
plan, 196, 197
plane, model of, 19b
projections, 196
Axes, co-ordinate, 150, 178
isometric, 444
metric, 443
trimetric, 443, 454
Axis, conjugate, 104
polar, 186
transverse, 104
Black thread used in plotting, 170
Board, drawing-, 3
Bow's notation, 514
Cam, heart-shaped, 144
Cams, 142
Cartridge paper, 3
Cast shadows, 406
Celluloid, 120
ruled, 10, 12
Centre of area, 530
of curvature, 90
of curvature of cycloid, 116
of curvature of epicycloid, 116
of curvature of hypocycloid, 116
of curvature of involute, 122
of curvature of roulette, 112
of gravity, 530
Chart, price, 167
Chisel-edged pencil, 4
Choice of scales, 168
Chord of angle, 19
< 'hords, scale of, 22
Circles and lines in contact, 50, 60-66
Circuital, 507
non-, 507
55
PRACTICAL GEOMETRY
Circular arc, length of, 112
Circular protractor, 3
Clinograph, 3
how to use, 5
Closing of force polygon, 510, 525,
532. 533
of link polygon, 525, 532, 533
side of vector polygon, 508
Co-latitude, 187
Compasses, 3
Complete quadrilateral, 34
Component motions, 130
Components, 504
rectangular, 504
Compounding vectors, 504
Concrete number, 486
Concurrent forces, 510
Conditions of equilibrium, 525
Cone and inscribed sphere, 320, 342
Cone, axis of, 70
double, 70, 71
generator of, 70
of indefinite length, 320
trace of, 322, 326
vertex of, 70
Cones, tangential properties of, 255
Conical surface, 544
Conic sections, 70
classification of, 70
definition of, 72
properties of, 72
Conjugate axis of hyperbola, 104
diameters of ellipse, 86
Connecting rod of engine, 138
Construction of ellipse, 76, 82, 83,
88, 96-98
of parabola, 99, 102
of regular polygons, 52
of triangles, 53-59
Contact of lines and circles, 50
of surfaces, 368
Continued product of lines, 490
Contour road-map, 167
Contours, 442, 472
Co-ordinates, axes of, 150, 178
of a point, 148-151, 178
origin of, 150, 178
polar, 149, 186
positive and negative, 151, 178
Co-ordinates, rectangular, 149, 178
r, 6, <p, 186
systems of, 149
Correction of errors of observation,
170
Cosine, definition of, 19
Cosines, table of, 20
Couples, 520, 523
Crane test, 170
jib, 517
Cube, 543
Curvature, 90, 91
centre of, 90, 91
circle of, 90, 91
radius of, 90, 91
Curve, envelope of, 120
equation to, 154
e volute of, 122
involute of, 122
logarithmic, 126
of sines, 128
parallel, 120, 144
plotting, 157
Curved surfaces, 308
Curves, French, 3
peculiarities of, 118
projection of, 308
special, no
tortuous, 384
Cusps, 118
Cutting, earthwork, 472
Cycloid, 114
construction of, 114
curtate, 114
normal to, 116
prolate, 114
tangent to, 116
Cycloidal curves, 114
special cases of, 117
Cylinder, right circular, 544
of indefinite length, 320
trace of, 324, 326
Cylindrical surface, 544
Definitions, 542-544
of similar polygons, 26
of sine, cosine, tangent, 18
Degree of accuracy in drawing, 2
Descriptive geometry, 176
INDEX
55i
Development of cone, 312
of octahedron, 198
of pyramid, 216
of sphere, 470
Diagonal scales, construction of, 14
Diagram, force, 518
of load-elongation, 166
of moments, 522, 526
Direct-acting steam engine, 138
Directed quantities, 502
Directrix, 72
Displacement of slide valve, 169
Displacements, triangle of, 399
Dividers, 3
Division, graphic, 487
of a given line, 10, 12
of line, internally and externally,
of polygon, 37
Dodecahedron, 462
Double cone, 70, 71
Draughtsman, 2
Draughtsmanship, good, 177
Drawing-board, 3
Drawing, errors in, 2
-paper, 3
to scale, 1
Earth's equator, 186
Earthwork, contoured, 472
Eccentricity, 72, 75, 105
Edge, straight-, 2
Efficiency-resistance curve, 171
Effort-resistance curve, 171
Elevation, front and side, 182
Ellipse, 70, 74
auxiliary circles of, 78
centre of, 74
conjugate diameters of, 86
construction of, 76, 82, 83, 88,
96, 97, 98
diameters of, 74
directrices of, 74
focal properties of, 92, 93, 94
foci of, 74, 76
major and minor axes of, 74, 76
mechanical method of describing
an, 76
mechanism for drawing an, 81
Ellipse, principal axes of, 74, 87
tangential properties of, 92, 93,
94
theorems on, 74, 75, 79, 80
Elongation-load diagram, 166
Embankments, contoured, 472
Engine-divided scale, 2
Envelope of a curve, 120
of a tangent, 478
Epicycloid, 114, 116
tangent to, 116
normal to, 116
Epitrochoid, 114
Equation to a curve, 154
to a straight line, 160
Equation, solution of cubic, 172
Equator, 186
Equiangular spiral, 126
Equilibrant, 510, 514
Equilibrium, 510
Errors in drawing, 2, 51
of observation, 170
Euclid, 1, 50, 176
Evolute of a curve, 122
Evolution, 494
Examination paper, 538
Experiment, plotting results of, 166
Extreme and mean ratio, 32
Figured plans, 198, 275
Focal properties of ellipse, 92, 93,
94
sphere, 72, 74, 322
Focus of a conic, 72
Force diagram, 518
scale, 522
Forces, concurrent, 510
moment of, 520, 526
parallel, 525, 526
polygon of, 510
Fourth proportional, 32, 488
Frame, hinged, 516, 518, 521
French curves, 3
Friction of crane, 170
Frustum of pyramid, 216
Fundamental rules, 193-197
Funicular polygon, 519, 524
Generator, 70
552
PRACTICAL GEOMETRY
Geometrical mean, 32
Geometry, analytical, 156
descriptive, 176
practical, 1
practical solid, 176
pure, 1
pure solid, 176
Glass paper, 4
Good draughtsmanship, 177
Graphic arithmetic, 486
statics, 502
Graphical solution of equations, 172
Gravity, centre of, 530
Harmonic division of a line, 34
mean, 34
motion, 128, 132
pencil, 34, 51
progression, 34
range, 34
Helical spring, 476
surface, 476
Helix, pitch angle of, 476
pitch of, 476
projection of, 476
Hinged frame, 516, 518, 521
Horizontal projection, 274
Hyperbola, 71, 104
asymptotes of, 104
centre of, 104
properties of, 104
rectangular, 154
tangent and normal to, 106
theorems on, 105
transverse axis of, 104
vertices of, 104
Hypocycloid, 114, 116
tangent and normal to, 116
Hypotrochoid, 114
Icosahedron, 463
Ill-conditioned constructions, 10
Important points, 385, 408
problem, 219
problem on shadows, 428
sections, 396
Inaccessible points, 30, 60
Inclinations of line, 188, 206
of plane, 191, 256
Inclined plane, 214
Inking-pen, 3
Inscribed sphere, 320, 342
Instantaneous centre, 88, 93, 112,
122, 141
Instruments, drawing, 1, 3
accuracy of, 4
setting of, 4
Integral powers, 494
Intercepts of line, 160
of plane, 190
Interpenetrations of solids, 384
Interpolation, 169
Intersection of cone and cylinder,
390-395
of cylinder and pyramid, 400
of line and plane, 220, 224, 234,
437
of prism and pyramid, 398
of shadows, 428
of sphere and prism, 396
of surfaces, 384
of surfaces of revolution, 402
of three planes, 250
of two cylinders, 388
of two planes, 250, 280
Involute, centre of curvature of, 122
of a curve, 122, 478
tangent and normal to, 122
Involution, 494
Isometric axes and scale, 444
projection, 444
scale, construction of, 446
Jib crane, 516
Laboratory test of a crane,' 170
Latitude, 187
Law of crane, 173
Length of circular arc, 112
Limits of accuracy, 2
Line and perpendicular plane, 204
and plane, intersection of, 220,
224, 234, 437
division of, 10
equation to straight, 160, 162
inclinations of, 206
of separation, 407, 416
possible positions of, 204
INDEX
55$
Line, powers of, 494, 500
projections of, 188
square root of, 496
to find equation to a, 162
traces of, 188, 208
unit, 486, 488
width of a, 2
Linear equation, 160
laws, 173
scale, 522
Lines, product of, 488
quotient of, 488
shortest distance between, 288
of slope, 224
Link motion, 139, 174
polygon, 519, 524
polygon, properties of, 524
polygon, closing of, 525, 532, 533
Loaded string, 518
Load-elongation diagram, 166
Locus, 66, 84
Logarithmic spiral, 124, 126
curve, 126, 500
Log of timber, volume of, 175
Longitude, 186
Major axis of an ellipse, 74
Map, contour road-, 167
Maps, contoured, 442
Mathematical instruments, 3
Maxima and minima, 169
Mean and extreme ratio, 32
geometrical, 32
harmonic, 34
ordinate, 44
proportional, 32
Measurements to scale, 177
Mechanical appliances, 3, 4
constraint, 136
method of describing ellipse, 76
Median, 56
Meridian circle, 186
plane, 186
Method of sections, 384
Metric axes, 443
directions, 443
lines, 443
plane, 443
projection, 442
Metric scales, 443
Mid-ordinate rule. 44
Minor axis of ellipse, 74
Miscellaneous problems, 462
Model of auxiliary plane, 196
of oblique plane, 190, 230
of perpendicular plane, 214
of planes of projection, 180, 193
ol ruled surface, 479
Models, use of, 177-191
Moment diagram, 522, 526
of a force, 520
of a system of forces, 525
partial, 525
scale, 522
Motions, component and resultant,
^a. S4
under mechanical constraint, 136
Multiplication, graphic, 487
Nodes, 118
Non-concurrent forces, 520, 524
resultant of, 524
Normal to any curve, 90
to a cone or cylinder, 548
to a cycloid, 116
to an ellipse, 92
to an epicycloid, 116
to a hyperbola, 106
to a hypocycloid, 116
to an involute, 122
to a parabola, 101
to a roulette, 112
Notation, in solid geometry, 182
Bow's, 514, 526
Number, concrete, 486
pure, 8, 486
Oblique plane, 190, 214, 230
converted into inclined plane,
232
model of, 190, 230
Observation, correction of errors of,
170
plotting results of, 166
Octahedron, 426
Order or sequence, 503
Ordinate, 44, 150
mean, 44
554
PRACTICAL GEOMETRY
Origin of co-ordinates, 150, 178
Orthographic projection, 182
of a conic, 78
Pantograph, 26
Paper,, drawing-, 3
Paper, squared, 152
Parabola, 71, 99
construction of, 99, 102
directrix of, 99
focus of, 99
normal to, 101
properties of, 100
tangent to, 101
theorems on, 100
Parallel curves, 120
forces, 525, 526
motion, 140
projection, 182, 458
rulers, 1
Parallelogram of displacements, 507
of forces, 511, 512
of vectors, 506
Partial moment, 525
resultant, 524
Pencil of rays, 34
quality of, 3
sharpening of, 4
Period of vibration, 128
Perpendicular plane, 214
model of, 214
Phase, 128
Plan, 182
Plane, 190
inclinations of, 191, 256
inclined, 214
intercepts of, 190
oblique, 190, 230
perpendicular, 214
rabatments of, 190, 219, 235,
244
scale of slope of, 224
through three points, 248
traces of, 190, 215, 231
Planes, angle between, 249, 284
front and side, 178
horizontal and vertical, 178
intersection of, 250, 280
of projection, 182, 193, 195
1 Planes of projection, models of, 180,
193
of reference, 178
Plotting, examples on, 172
a straight line, 162
curves, 157
points, 150
results of experiment, 166
results of observation, 166
Point in space, 178, 180
of inflexion, 118
size of a, 2
(x,y), 144
(x, y, 2), 179
Polar axis, 186
co-ordinates, 149, 186
triangle, 466
Pole, 186, 519
of spiral, 124
Polygon, area of, 36, 492
division of, 37
funicular, 519
link, 519
of displacements, 508
of forces, 510
vector, 508
Polygons, construction of regular, 52
similar, 26
Polyhedron, 462
sphere circumscribed about, 464
sphere inscribed in, 464
Position in space, 176
in space defined and exhibited,
178, 180, 292
of a point, 178, 180
of a point in a plane, 148, 150
Powers of a line, 494, 500
Practical geometry, 1
solid geometry, 176
Pressure of steam, 175
] Price chart, 167
I Pricker, how to make a, 3
j Problems, miscellaneous, 462
Product of lines, 488
Projection, definitions relating to,
182
isometric, 444
horizontal, 274
metric, 442, 458
INDEX
555
Projection, oblique, 182, 458
of cone, 320
of curves, 308
of cylinder, 320
of a helix, 476
of surface of revolution, 334
of V-threaded screw, 476
orthogonal or orthographic, 182
parallel, 182
planes of, 182, 193, 195
sectional, 198
trimetric, 454, 458
Projections, figured, 198
Projective properties of ellipse, 78
Projectors, 182, 196
Properties of conic sections, 72
of cone and cylinder, 320
Proportional, fourth, 32, 488
mean, 32
third, 32
Proportion and ratio, 8
Protractor, circular, 3
Pure geometry, 1
number, 8, 486
solid geometry, 176
Pyramid, development of, 216
frustum of, 216
Quadrilateral, complete, 34
Quantity, directed, 502
(r, 6, <p) co-ordinated, 186
Rabatment, 190, 206, 218, 244
model to show, 196, 245
of a point, 196, 244
Radial projection of a conic, 78
Radius vector, 124
Range, harmonic, 34
Rankine's construction for length of
circular arc, 112
Ratio and proportion, 8
representation by symbols, 8, 9
theorems on, 8, 9
Ratio, extreme and mean, 32
Ratios, definition of trigonometrical,
18, 19
Rays, divergent, 407
parallel, 407
pencil of, 34
Rectangular components of a vector,
504. 512
co-ordinates, 149, 150, 178
hyperbola, 154
Regular polyhedra, 462, 543
Representation of vectors, 503
Resistance-efficiency curve, 171
-effort curve, 171
Resolving vectors, 504, 512
Resultant, 504, 512, 524
motions, 130
of non-concurrent forces, 524
partial, 524
Results of experiments, plotting of,
166
of observation, plotting of, 166
Revolution, surface of, 334, 336,
342, 422, 544
Right-angled triangle, solution of,
16
Road-map, 167
Rolling curves, no
Roulettes, 1 10
base of, 1 10
centre of curvature of, 112
construction of, no
definition of, no
normal to, 112
Rule for area, 44
mid-ordinate, 44
ordinary, 44
Simpson's, 44, 45
Weddle's, 45
Ruled celluloid, 10, 12
surface, model of, 479
surface, section of, 478
tracing-paper, 10
Rulers, parallel, 1
Rules, fundamental, 193
for adding vectors, 507, 508
for mean ordinate, 44
Scale, r, 6
choice of, 168
close and open divided, 6
decimally divided, 6
description of, 6
diagonal, 14
engine-divided, 2
556
PRACTICAL GEOMETRY
Scale, examples on use of, 7
force, 522
isometric, 444, 446
linear, 522
metric, 443, 458
moment, 522
of chords, 22
of slope of plane, 224
of speed, 12
trimetric, 443, 454
Screw thread, 476
pitch of, 476
V-threaded, 476
Sealing-wax, use of, 3
Section I., plane geometry, 1
II., solid geometry, 176
III., graphics, 486
of sphere by plane, 386
Sectional projections, 198
Sections, conic, 70
method of, 384
Separation, line ot, 407, 416**
Sequence, or order, 503
Set-squares, 1, 3
adjustment of, 4
errors in, 2
Setting of instruments, 4
Shadows of bolt, 424
cast, 406
of circle, line, point, 409-411
geometrical, 406
important problem on, 428
intersection of, 428
of rivet, 426
of simple solids, 411-423
theorems on, 407, 408, 424
Sharpening of pencil, 4
Shortest distance between two lines,
212, 288
Similar polygons, 26
Simple harmonic motion, 128, 132
amplitude of, 128
period of, 128
phase of, 128
Sine curve, 128, 476
definition of, 18
Sines, table of, 20, 21
Size of a point, 2
Skeleton diagrams, 137
Slide valve, 139, 169
Slope, scale of, 224, 275
line of, 224
Solids in given positions, 292
Solution of equations, 172
of right-angled triangle, 16
Space of three dimensions, 177
Special curves, no
points, 385
Sphere inscribed in cone, 320
in cylinder, 320
section by plane, 386
Spherical triangles, 466
Spiral curves, 124
equiangular, 126
logarithmic, 124, 126
of Archimedes, 124
springs, 476
Squared paper, 152
Square root of line, 496
of number, 496
Statics, graphic, 502
Steam pressure, 175
Stephenson's link motion, 174
Straight-edge, 2
line, average position of, 170
line, equation to,
line, plotting of, 162
line, projections of, 188, 205
String, loaded, 518
Subtraction, graphic, 487
Summation, graphic, 487
Surfaces, development of, 198,
216, 470
in contact, 368
revolution, 334, 336, 342, 422
ruled, 478
System, concurrent, 510
Systems of co-ordin?tes, 149, 180
Tables of sines, cosines, etc., 20, 21
Tangent, definition of, 19
planes to surfaces, 340
to any curve, 90
to cycloid, 116
to draw a, 50
to ellipse, 92
to epicycloid, 116
to hyperbola, 106
INDEX
557
Tangent to hypocycloid, 116
to involute, 122
to parabola, 101
Tangential properties of cones, 255
properties of ellipse, 92, 93, 94
Tangents, table of, 20, 21
Tee- square, 3
Teeth, wheel, no, 120
Temperature of steam, 175
Templates, 3, in, 120, 138
ruled, 10
Testing machine, 166
Test of crane, 170
Tetrahedron, 464, 543
height of, 302
Theorems, 544
on couples, 520
on the ellipse, 74, 75, 79, 80,
92
on forces, 511, 524, 525
on the hyperbola, 105
on involutes and evolutes, 122
on the line and circle, 51
on metric projection, 459
on the parabola, 100
on shadows, 407, 408, 424
on similar polygons, 26, 27, 42
Third proportional, 32
Thread used in plotting, 170
screw, 476
Three planes of reference, 178
co-ordinate planes, 178
metric axes, 443, 458
Tie of crane, 516
Tools, 2
Tortuous curves, 384
Trace of cone, 322, 326
of cylinder, 324, 326
Traces of a line, 188, 208
of a plane, 190, 214, 231
Tracing-paper, 170
ruled, 10, 12
template, 3, in, 120, 138
Trammel for ellipse, 82, 83, 87
triangular, 85, 87, 88
Transversal, 34
Triangle of displacements, 506
of forces, 511, 513
of vectors, 506
polar, 466
rabatment of, 190, 219
spherical, 466
solution of right-angled, 16
Triangles, construction of, 52-59
Triangular trammel, 85, 87, 88
Trigonometrical tables, 20, 21
Trihedral angles, 179, 466
Trimetric axes, 443, 454, 458
scales, 454, 458
True alignment, 2
Trying plane, 2, 4
! Unit line, 486, 488
Useful construction, 243
Use o*f models, 177, 191
of squared paper, 152
Vector, definition of, 502
equality, 508, 512
parallelogram, 506
polygon, 508
radius, 124
representation of, 503
summation, 508
triangle, 506
Vectorally equal, 508, 512
Vectorial angle, 124
Vertical plane, 214
Volume of log of timber, 175
of 1 lb. of steam, 175
V-threaded screw, 476
Watt's parallel motion, 140
Ways of defining angles, 18
Wheel teeth, no, 120
Width of a line, 2
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QA
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